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## [$AcWing$ $1134$. 最短路计数](https://www.acwing.com/problem/content/1136/)
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### 一、题目描述
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给出一个 $N$ 个顶点 $M$ 条边的 **无向无权图**，顶点编号为 $1$ 到 $N$。
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问从顶点 $1$ 开始，到其他每个点的 **最短路有几条**。
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**输入格式**
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第一行包含 $2$ 个正整数 $N,M$，为图的顶点数与边数。
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接下来 $M$ 行，每行两个正整数 $x,y$，表示有一条顶点 $x$ 连向顶点 $y$ 的边，**请注意可能有自环与重边**。
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**输出格式**
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输出 $N$ 行，每行一个非负整数，第 $i$ 行输出从顶点 $1$ 到顶点 $i$ 有多少条不同的最短路，由于答案有可能会很大，你只需要输出对 $100003$ 取模后的结果即可。
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如果无法到达顶点 $i$ 则输出 $0$。
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**数据范围**
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$1≤N≤105,1≤M≤2×10^5$
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**输入样例**：
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```cpp {.line-numbers}
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5 7
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1 2
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1 3
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2 4
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3 4
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2 3
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4 5
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4 5
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```
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**输出样例**：
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```cpp {.line-numbers}
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1
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1
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1
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2
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4
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```
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### 二、解题思路
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> 要求 **最短路计数** 要满足: **不能存在路径和为$0$的环**，因为存在的话某个环中不断的转圈不出来，而路径长度不变,那么被更新的点的条数就为$INF$了
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#### 1. 更新记录最短路的数量
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#### 2. 求最短路的算法
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* $bfs$
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只入队一次，出队一次, **可以抽象成拓扑图**， 因为它可以保证被更新的点的父节点一定已经是最短距离了，并且这个点的条数已经被完全更新过了。
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* $dijkstra$
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每个点只出队一次, **可以抽象成拓扑图**， 同理由于每一个出队的点 **一定** 已经是 **最短距离**，并且它出队的时候是队列中距离最小的点，这就代表他的最短距离条数已经被完全更新了，所以构成拓扑性质。
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<!-- 让表格居中显示的风格 -->
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<style>
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.center
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{
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width: auto;
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display: table;
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margin-left: auto;
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margin-right: auto;
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}
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</style>
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<div class="center">
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| 题型 | 方法 | 可行性 | 备注 |
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| ----------------------------------- | -------------------------------- | ------ | --------------------------------------------- |
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| 最短路+最短路数量 | $Dijkstra+Heap+DP$ | $√$ | <font color='red' size='4'><b>推荐</b></font> |
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| | $bfs+DP$ | $√$ | 了解 |
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| | $spfa+DP$ | $√$ | 了解 |
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| | 求最短路+建拓扑图+遍历拓扑图$DP$ | $√$ | <font color='blue' size=4><b> 学习</b></font> |
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| 最短路+最短中数量+次短路+次短路数量 | $Dijkstra$堆优化+$DP$+拆点 | $√$ | <font color='red' size=4><b>推荐</b></font> |
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| | $bfs+DP$ | $×$ | |
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| | $spfa+DP$ | $×$ | |
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| | 求最短路+建拓扑图+遍历拓扑图$DP$ | $√$ | <font color='blue' size=4><b>学习</b></font> |
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</div>
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#### $bfs$解法
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 4e5 + 10;
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const int MOD = 100003;
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int h[N], ne[N], e[N], idx;
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void add(int a, int b) {
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e[idx] = b, ne[idx] = h[a], h[a] = idx++;
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}
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int cnt[N]; //从顶点1开始，到其他每个点的最短路有几条
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int dist[N]; //最短距离
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int n, m;
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void bfs() {
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memset(dist, 0x3f, sizeof dist);
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queue<int> q;
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q.push(1);
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cnt[1] = 1; //从顶点1开始，到顶点1的最短路有1条
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dist[1] = 0; //距离为0
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while (q.size()) {
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int u = q.front();
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q.pop();
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (dist[j] > dist[u] + 1) {
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dist[j] = dist[u] + 1;
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q.push(j);
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cnt[j] = cnt[u];
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} else if (dist[j] == dist[u] + 1)
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cnt[j] = (cnt[j] + cnt[u]) % MOD;
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}
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}
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}
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int main() {
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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while (m--) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(a, b), add(b, a);
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}
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bfs();
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for (int i = 1; i <= n; i++) printf("%d\n", cnt[i]);
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return 0;
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}
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```
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#### $Dijkstra$解法
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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typedef pair<int, int> PII;
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const int N = 100010, M = 400010;
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const int MOD = 100003;
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int n, m;
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int cnt[N];
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int dist[N];
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bool st[N];
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int h[N], e[M], ne[M], idx;
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void add(int a, int b) {
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e[idx] = b, ne[idx] = h[a], h[a] = idx++;
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}
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void dijkstra() {
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memset(dist, 0x3f, sizeof dist);
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dist[1] = 0;
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// 出发点到自己的最短路径只能有1条
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cnt[1] = 1;
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// 小顶堆q
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priority_queue<PII, vector<PII>, greater<PII>> pq;
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pq.push({0, 1});
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while (pq.size()) {
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auto t = pq.top();
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pq.pop();
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int u = t.second, d = t.first;
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if (st[u]) continue;
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st[u] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (dist[j] > d + 1) {
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dist[j] = d + 1;
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cnt[j] = cnt[u];
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pq.push({dist[j], j});
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} else if (dist[j] == d + 1)
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cnt[j] = (cnt[j] + cnt[u]) % MOD;
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}
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}
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}
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int main() {
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scanf("%d %d", &n, &m);
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memset(h, -1, sizeof h);
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while (m--) {
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int a, b;
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scanf("%d %d", &a, &b);
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add(a, b), add(b, a);
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}
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dijkstra();
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for (int i = 1; i <= n; i++) printf("%d\n", cnt[i]);
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return 0;
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}
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```
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### 三、扩展练习
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[$POJ$ - $3463$ $Sightseeing$](http://poj.org/problem?id=3463)
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**最短路计数+ 次短路计数**
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**题意**
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已知一张图（单向边），起点$S$和终点$F$，求从 $S $到 $F$ 的 **最短路** 和 **比最短路长$1$** 的路径的 **条数之和**。
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**分析**
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只需要计算出最短路的条数和距离、次短路的距离和条数，最后判断最短路和次短路的关系即可，在$dijkstra$求最短路的基础上，**加一维** 保存从起点到该点的 **最短路** 和 **次短路**，**同时记录相应的数量**
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如果当前点的最短路或次短路更新了，那么这个点可能松弛其他点，加入优先队列；如果等于最短路或次短路，相应的数量就加
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关于代码中①②的自我解释：
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①：一个点可能被几个点松弛，每松弛一次就入一次队列，那么这个点也会出几次队列，也就会重复松弛后面的点，路径数量就会算重，所以要标记每个点的最短路和次短路是否出队，所以入队时要记录入队的是最短路还是次短路，再者，这个标记本来具有优化作用，这里求数量又多了一层意义，就不能使用其他的优化方式了
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②：为什么当前点的次短路更新为当前点的最短路时，还要入队呢？当前点的最短路被松弛时已经入过队了啊，原因很简单，因为每个点入队时都记录了入队的是最短路，还是次短路，再一次更新最短路时，数量也随之更新，而上一次入队的最短路的数量就不再是自己的数量了，而是更新后最短路的数量
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```cpp {.line-numbers}
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#include <queue>
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#include <cstring>
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#include <iostream>
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#include <algorithm>
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using namespace std;
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#define x first
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#define y second
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const int N = 1e3 + 13;
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const int M = 1e6 + 10;
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int n, m, u, v, s, f;
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int dist[N][2], cnt[N][2];
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bool st[N][2];
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//链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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struct Node {
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// u: 节点号
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// d:目前结点v的路径长度
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// k:是最短路0还是次短路1
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int u, d, k;
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// POJ中结构体，没有构造函数，直接报编译错误
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Node(int u, int d, int k) {
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this->u = u, this->d = d, this->k = k;
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}
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const bool operator<(Node x) const {
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return d > x.d;
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}
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};
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void dijkrsta() {
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priority_queue<Node> q; //通过定义结构体小于号，实现小顶堆
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memset(dist, 0x3f, sizeof(dist)); //清空最小距离与次小距离数组
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memset(cnt, 0, sizeof(cnt)); //清空最小距离路线个数与次小距离路线个数数组
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memset(st, 0, sizeof(st)); //清空是否出队过数组
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cnt[s][0] = 1; //起点s，0:最短路，1:有一条
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cnt[s][1] = 0; //次短路，路线数为0
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dist[s][0] = 0; //最短路从s出发到s的距离是0
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dist[s][1] = 0; //次短路从s出发到s的距离是0
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q.push(Node(s, 0, 0)); //入队列
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while (q.size()) {
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Node x = q.top();
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q.pop();
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int u = x.u, k = x.k, d = x.d;
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if (st[u][k]) continue; //①
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st[u][k] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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int dj = d + w[i]; //原长度+到节点j的边长
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if (dj == dist[j][0]) //与到j的最短长度相等，则更新路径数量
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cnt[j][0] += cnt[u][k];
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else if (dj < dist[j][0]) { //找到更小的路线，需要更新
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dist[j][1] = dist[j][0]; //次短距离被最短距离覆盖
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cnt[j][1] = cnt[j][0]; //次短个数被最短个数覆盖
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dist[j][0] = dj; //更新最短距离
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cnt[j][0] = cnt[u][k]; //更新最短个数
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q.push(Node(j, dist[j][1], 1)); //②
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q.push(Node(j, dist[j][0], 0));
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} else if (dj == dist[j][1]) //如果等于次短
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cnt[j][1] += cnt[u][k]; //更新次短的方案数，累加
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else if (dj < dist[j][1]) { //如果大于最短，小于次短，两者中间
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dist[j][1] = dj; //更新次短距离
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cnt[j][1] = cnt[u][k]; //更新次短方案数
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q.push(Node(j, dist[j][1], 1)); //次短入队列
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}
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}
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}
|
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}
|
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int main() {
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int T;
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scanf("%d", &T);
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while (T--) {
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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while (m--) {
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int a, b, c;
|
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scanf("%d %d %d", &a, &b, &c);
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add(a, b, c);
|
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|
}
|
|
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|
|
//起点和终点
|
|
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|
|
scanf("%d %d", &s, &f);
|
|
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//计算最短路
|
|
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|
|
dijkrsta();
|
|
|
|
|
|
//输出
|
|
|
|
|
|
printf("%d\n", cnt[f][0] + (dist[f][1] == dist[f][0] + 1 ? cnt[f][1] : 0));
|
|
|
|
|
|
}
|
|
|
|
|
|
return 0;
|
|
|
|
|
|
}
|
|
|
|
|
|
```
|
