费用流 hdu1853 Cyclic Tour

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;}