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## [$POJ3216$ $Repairing$ $Company$](http://poj.org/problem?id=3216)
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### 一、题目描述
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有$n$个维修站，给出了一个邻接矩阵(对称阵)表示每个维修站到其他维修站的花费的时间，$-1$表示不可达，然后给出了$m$个任务，给出了每个任务要在哪个维修站进行，起始时间 和 任务花费时间，问至少要几个维修人员才能准时进行任务。
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### 二、题目分析
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很明显的最小路径覆盖问题，刚开始脑子抽了，没求最短路直接就做了，题目只给了两点间直接到达的时间，还可以间接到达，用$floyd$求出最短路。。。
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### $Code$
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```cpp {.line-numbers}
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#include <iostream>
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#include <algorithm>
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#include <queue>
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#include <map>
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#include <cstring>
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#include <vector>
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#include <stack>
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#include <cstdio>
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using namespace std;
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const int N = 25;
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const int M = 205;
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const int INF = 0x3f3f3f3f;
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struct Task {
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int id; // 维修站
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int cost; // 花费时间
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int st; // 起始时间
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} task[M];
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int m, n, g[N][N];
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// 匈牙利算法
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int match[M], st[M];
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bool dfs(int u) {
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for (int i = 0; i < m; i++) {
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if (st[i]) continue;
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// 这里很妙，不是真的把图建出来，而是直接利用原来的邻接矩阵，通过条件判断来决策，减少了代码!
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if (task[u].st + task[u].cost + g[task[u].id][task[i].id] > task[i].st) continue;
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st[i] = 1;
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if (match[i] == -1 || dfs(match[i])) {
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match[i] = u;
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return 1;
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}
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}
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return 0;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ3216.in", "r", stdin);
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#endif
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while (~scanf("%d%d", &n, &m) && n + m) {
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// 题目中给出的不可达值为-1，因为要求最短路，所以设置为INF
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++) {
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scanf("%d", &g[i][j]);
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if (g[i][j] == -1) g[i][j] = INF;
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}
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// Floyd求任意两点间最短距离
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for (int k = 1; k <= n; k++)
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++)
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g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
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// m个任务
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for (int i = 0; i < m; i++)
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scanf("%d%d%d", &task[i].id, &task[i].st, &task[i].cost);
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// Hungary
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memset(match, -1, sizeof match);
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int res = 0;
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for (int i = 0; i < m; i++) { // 枚举所个任务
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memset(st, 0, sizeof st);
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if (dfs(i)) res++;
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}
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// 输出结果
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printf("%d\n", m - res);
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}
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return 0;
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}
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```
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