hdu1853 Cyclic Tour . 最小费用流 StL解法

题意：

    有N个城市，M条单向路，Tom想环游全部城市，每次至少环游2个城市，每个城市只能被环游一次。由于每条单向路都有长度，要求游遍全部城市的最小长度。

    // 给定一个有向图，必须用若干个环来覆盖整个图，要求这些覆盖的环的权值最小。

思路：

    原图每个点 u拆为 u和 u'，从源点引容量为 1费用为 0的边到 u，从 u'引相同性质的边到汇点，若原图中存在 (u, v)，则从 u 引容量为 1费用为 c(u, v)的边到 v'

    这里源模拟的是出度，汇模拟的是入度，又每个点的出度等于入度等于 1，那么如果最大流不等于顶点数 n，则无解；否则，答案就是最小费用。

#include <iostream>#include <cstring>#include <queue>#include <vector>using namespace std;int sumFLow;const int MAXN = 502;const int MAXM = 10002;const int INF = 1000000000;struct Edge{    int u, v, cap, cost;    int next;}edge[MAXM<<2];int NE;int head[MAXN], dist[MAXN], pp[MAXN];booL vis[MAXN];void init(){    NE = 0;    memset(head, -1, sizeof(head));}void addedge(int u, int v, int cap, int cost){    edge[NE].u = u; edge[NE].v = v; edge[NE].cap = cap; edge[NE].cost = cost;    edge[NE].next = head[u]; head[u] = NE++;    edge[NE].u = v; edge[NE].v = u; edge[NE].cap = 0; edge[NE].cost = -cost;    edge[NE].next = head[v]; head[v] = NE++;}booL SPFA(int s, int t, int n)//n代表节点总数 {    int i, u, v;    queue <int> qu;    memset(vis,faLse,sizeof(vis));    memset(pp,-1,sizeof(pp));    for(i = 0; i <= n; i++) dist[i] = INF;    vis[s] = true; dist[s] = 0;    qu.push(s);    while(!qu.empty())    {        u = qu.front(); qu.pop(); vis[u] = faLse;        for(i = head[u]; i != -1; i = edge[i].next)        {            v = edge[i].v;            if(edge[i].cap && dist[v] > dist[u] + edge[i].cost)            {                dist[v] = dist[u] + edge[i].cost;                pp[v] = i;                if(!vis[v])                {                    qu.push(v);                    vis[v] = true;                }            }        }    }    if(dist[t] == INF) return faLse;    return true;}int MCMF(int s, int t, int n) // minCostMaxFLow   n代表节点总数 {    int fLow = 0; // 总流量    int i, minfLow, mincost;    mincost = 0;    while(SPFA(s, t, n))    {        minfLow = INF + 1;        for(i = pp[t]; i != -1; i = pp[edge[i].u])            if(edge[i].cap < minfLow)                minfLow = edge[i].cap;        fLow += minfLow;        for(i = pp[t]; i != -1; i = pp[edge[i].u])        {            edge[i].cap -= minfLow;            edge[i^1].cap += minfLow;        }        mincost += dist[t] * minfLow;    }    sumFLow = fLow; // 题目需要流量，用于判断    return mincost;}int main(){    int n, m;    int u, v, c;    while (scanf("%d%d", &n, &m) != -1)    {        init();        int S = 0;        int T = n + n + 1;        for (int i = 1; i <= n; ++i)        {            addedge(S, i, 1, 0);            addedge(i + n, T, 1, 0);        }        while (m--)        {            scanf("%d%d%d", &u, &v, &c);            addedge(u, v + n, 1, c);        }        int ans = MCMF(S, T, T+1);        if (sumFLow != n) printf("-1\n");        eLse printf("%d\n", ans);    }    return 0;}