传送门：点击打开链接
题意：给n个点和M条有向边，要找出许多个环出来，每个环点至少有2个，所有的点都要被环覆盖1次，且只能为1次。问所有环的长度之和
这题也可以用KM来做，这里主要是练习费用流的建图
对于这题，建图也是非常的奇妙的
由于每个点的入度都是1，出度都是1
所以会想到把每个点拆分成2个点，用i和i+n来表示
然后将源点与所有的i连接起来，将汇点与所有的i+n连接起来，容量都是1
对于每一条边(u,v)，添加边(u,v+n,1,cost)，让第一层的点连接到第二层去
其实就是将点拆分成两层，一层是输入层，一层是输出层。（脑补一下..）
最后检查最大流是否等于n，如果不等于就无解，否则就输出最小费用
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cctype>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)using namespace std;
typedef long long LL;
typedef pair<int, int> PII;const int MX = 1e3 + 5;
const int MM = 2e5 + 5;
const int INF = 0x3f3f3f3f;struct Edge {int to, next, cap, flow, cost;Edge() {}Edge(int _to, int _next, int _cap, int _flow, int _cost) {to = _to; next = _next; cap = _cap; flow = _flow; cost = _cost;}
} E[MM];int Head[MX], tol;
int pre[MX]; //储存前驱顶点
int dis[MX]; //储存到源点s的距离
bool vis[MX];
int N;//节点总个数，节点编号从0~N-1void init(int n) {tol = 0;N = n + 2;memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v, int cap, int cost) {E[tol] = Edge(v, Head[u], cap, 0, cost);Head[u] = tol++;E[tol] = Edge(u, Head[v], 0, 0, -cost);Head[v] = tol++;
}
bool spfa(int s, int t) {queue<int>q;for (int i = 0; i < N; i++) {dis[i] = INF;vis[i] = false;pre[i] = -1;}dis[s] = 0;vis[s] = true;q.push(s);while (!q.empty()) {int u = q.front();q.pop();vis[u] = false;for (int i = Head[u]; i != -1; i = E[i].next) {int v = E[i].to;if (E[i].cap > E[i].flow && dis[v] > dis[u] + E[i].cost) {dis[v] = dis[u] + E[i].cost;pre[v] = i;if (!vis[v]) {vis[v] = true;q.push(v);}}}}if (pre[t] == -1) return false;else return true;
}//返回的是最大流， cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost) {int flow = 0;cost = 0;while (spfa(s, t)) {int Min = INF;for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {if (Min > E[i].cap - E[i].flow)Min = E[i].cap - E[i].flow;}for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {E[i].flow += Min;E[i ^ 1].flow -= Min;cost += E[i].cost * Min;}flow += Min;}return flow;
}inline int read() {char c = getchar();while(!isdigit(c)) c = getchar();int x = 0;while(isdigit(c)) {x = x * 10 + c - '0';c = getchar();}return x;
}int main() {int n, m; //FIN;while(~scanf("%d%d", &n, &m)) {int s = 0, t = 2 * n + 1;init(t);for(int i = 1; i <= n; i++) {edge_add(s, i, 1, 0);edge_add(i + n, t, 1, 0);}for(int i = 1; i <= m; i++) {int u, v, cost;u = read(); v = read(); cost = read();edge_add(u, n + v, 1, cost);}int ans = 0;if(minCostMaxflow(s, t, ans)!=n) printf("-1\n");else printf("%d\n", ans);}return 0;
}