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@ -46,52 +46,93 @@ $1≤n≤100 ,0≤k≤200,1≤f(i,j)≤1000$
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$kruskal$算法是 **求连通块** 的，所以这个题直接用 $kruskal$ 很容易求出来。
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```cpp {.line-numbers}
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if (cnt < n - 1) res = INF;
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```
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这句话需要注释掉，比如下面的数据用例：
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```cpp {.line-numbers}
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6 6
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1 2 5
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1 3 4
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2 3 8
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4 5 7
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4 6 2
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5 6 1
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```
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我们发现，$1,2,3$是一伙，$4,5,6$是另一伙，这两个家庭不通！如果按照模板的意思，那么就没有最小生成树！
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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 = 110, M = 210;
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int n, m, fa[N];
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const int INF = 0x3f3f3f3f;
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//结构体
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struct Edge {
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int a, b, w;
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bool operator<(const Edge &t) {
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return w < t.w;
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int n, m; // n条顶点,m条边
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int res; // 最小生成树的权值和
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int cnt; // 最小生成树的结点数
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// Kruskal用到的结构体
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struct Node {
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int a, b, c;
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bool const operator<(const Node &t) const {
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return c < t.c; // 边权小的在前
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}
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} e[M];
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} edge[M]; // 数组长度为是边数
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//并查集
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// 并查集
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int p[N];
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int find(int x) {
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if (fa[x] != x) fa[x] = find(fa[x]); //路径压缩
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return fa[x];
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if (p[x] != x) p[x] = find(p[x]);
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return p[x];
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}
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// Kruskal算法
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void kruskal() {
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// 1、按边权由小到大排序
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sort(edge, edge + m);
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// 2、并查集初始化
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for (int i = 1; i <= n; i++) p[i] = i;
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// 3、迭代m次
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for (int i = 0; i < m; i++) {
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int a = edge[i].a, b = edge[i].b, c = edge[i].c;
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a = find(a), b = find(b);
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if (a != b)
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p[a] = b, res += c, cnt++; // cnt是指已经连接上边的数量
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}
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// 这句话需要注释掉，原因如下：
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/*
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6 6
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1 2 5
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1 3 4
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2 3 8
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4 5 7
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4 6 2
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5 6 1
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我们发现，1,2,3是一伙，4,5,6是另一伙，这两个家庭不通！如果按照模板的意思，那么就没有最小生成树！
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这么说是没有问题的，但本题不是求最小生成树，而是求最小生成森林！所以，下面的特判需要注释掉！
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*/
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// 4、特判是不是不连通
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// if (cnt < n - 1) res = INF;
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}
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int main() {
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cin >> n >> m;
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//并查集初始化
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for (int i = 1; i <= n; i++) fa[i] = i;
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int sum = 0;
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// Kruskal算法直接记录结构体
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for (int i = 0; i < m; i++) {
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int a, b, c;
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cin >> a >> b >> c;
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e[i] = {a, b, c};
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edge[i] = {a, b, c};
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sum += c;
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}
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sort(e, e + m); //不要忘记e数组的长度是边的数量
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int res = 0;
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//枚举每条边
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for (int i = 0; i < m; i++) {
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int a = find(e[i].a), b = find(e[i].b), c = e[i].w;
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if (a != b)
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fa[a] = b;
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else
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res += c; //去掉的边权
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}
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printf("%d\n", res);
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kruskal();
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printf("%d\n", sum - res);
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return 0;
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}
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```
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### 三、$Prim$算法
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@ -103,39 +144,46 @@ using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 110;
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int w[N][N];
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int dist[N];
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bool st[N];
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int n, m, sum;
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int b[N];//桶，记录哪些点已经处理过了，找出未处理过的进行一下Flood Fill
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int prim(int source) {
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memset(dist, 0x3f, sizeof dist);
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dist[source] = 0;
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b[source] = 1;
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int res = 0;
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for (int i = 1; i <= n; i++) {
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int b[N];
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int n, m;
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int g[N][N]; // 稠密图，邻接矩阵
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int dis[N]; // 这个点到集合的距离
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bool st[N]; // 是不是已经使用过
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int res; // 最小生成树里面边的长度之和
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int sum; // 总边长
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// 普利姆算法求最小生成树
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int prim(int s) {
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// 由于调用多次prim，所以每次需要清零
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memset(dis, 0x3f, sizeof dis);
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res = 0;
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// 标识
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b[s] = 1;
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for (int i = 0; i < n; i++) { // 迭代n次
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int t = -1;
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for (int j = 1; j <= n; j++)
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if (!st[j] && (t == -1 || dist[t] > dist[j]))
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t = j;
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st[t] = true;
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if (!st[j] && (t == -1 || dis[t] > dis[j])) t = j;
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if (dist[t] != INF) res += dist[t], b[t] = 1;
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// if (i && dis[t] == INF) return INF; // 非连通图，没有最小生成树
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if (i && dis[t] != INF) res += dis[t], b[t] = 1;
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for (int j = 1; j <= n; j++)
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dist[j] = min(dist[j], w[t][j]);
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if (!st[j] && g[t][j] < dis[j]) dis[j] = g[t][j];
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st[t] = true;
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}
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return res;
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}
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int main() {
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cin >> n >> m;
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memset(w, 0x3f, sizeof w);
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memset(g, 0x3f, sizeof g);
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while (m--) {
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int a, b, c;
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cin >> a >> b >> c;
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w[a][b] = w[b][a] = c;
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g[a][b] = g[b][a] = c;
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sum += c; // 总边长
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}
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@ -146,4 +194,5 @@ int main() {
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printf("%d\n", sum - s);
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return 0;
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}
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```
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