poj 2526 Minimum Cost【最小费用最大流】

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题目链接：http://poj.org/problem?id=2516

题意：
n个店主 m个供应商 k种货物给你店主对k种货物的需求及供货商k种货物的囤货量及K种运输费用。

解法：k次费用流，分别求每种货物的费用。源点到供应点建边，店主到汇点建边，费用均为0，容量为1。然后供应点到店主建边，费用为矩阵，容量无穷大即可。

代码：

/*POJ 2195 Going Home邻接矩阵形式最小费用最大流*/#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<queue>using namespace std;//************************************************************//最小费用最大流算法//SPFA求最短路//邻接矩阵形式//初始化:cap:容量，没有边为0//cost:耗费，对称形式，没有边的也为0//c是最小费用//f是最大流//*******************************************************const int MAXN = 500;const int INF = 0x3fffffff;int cap[MAXN][MAXN];//容量，没有边为0int flow[MAXN][MAXN];//耗费矩阵是对称的，有i到j的费用，则j到i的费用为其相反数int cost[MAXN][MAXN];//花费int n;//顶点数目0~n-1int f;//最大流int c;//最小费用int start, End;//源点和汇点bool vis[MAXN];//在队列标志int que[MAXN];int pre[MAXN];int dist[MAXN];//s-t路径最小耗费bool SPFA(){    int front = 0, rear = 0;    for (int u = 0; u <= n; u++)    {        if (u == start)        {            que[rear++] = u;            dist[u] = 0;            vis[u] = true;        }        else        {            dist[u] = INF;            vis[u] = false;        }    }    while (front != rear)    {        int u = que[front++];        vis[u] = false;        if (front >= MAXN)front = 0;        for (int v = 0; v <= n; v++)        {            if (cap[u][v]>flow[u][v] && dist[v]>dist[u] + cost[u][v])            {                dist[v] = dist[u] + cost[u][v];                pre[v] = u;                if (!vis[v])                {                    vis[v] = true;                    que[rear++] = v;                    if (rear >= MAXN)rear = 0;                }            }        }    }    if (dist[End] >= INF)return false;    return true;}void minCostMaxflow(){    memset(flow, 0, sizeof(flow));    c = f = 0;    while (SPFA())    {        int Min = INF;        for (int u = End; u != start; u = pre[u])            Min = min(Min, cap[pre[u]][u] - flow[pre[u]][u]);        for (int u = End; u != start; u = pre[u])        {            flow[pre[u]][u] += Min;            flow[u][pre[u]] -= Min;        }        c += dist[End] * Min;        f += Min;    }}//************************************************************int tmp;int a[10000][55];int b[10000][55];int main(){    int N, M, K;    while (~scanf("%d%d%d", &N, &M, &K))    {        if (N == 0 && M == 0 && K == 0) break;        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        start = 0;        n = N + M+ 2;        End = M + N + 1;        int need = 0;        for (int i = 1; i <= N; i++)        {            for (int k = 1; k <= K;k++)            {                scanf("%d",&a[i][k]);                need += a[i][k];            }        }        for (int i = 1; i <= M; i++)        {            for (int k = 1; k <= K; k++)            {                scanf("%d", &b[i][k]);            }        }        int ans = 0;        int res = 0;        for (int kk = 1; kk <= K; kk++)        {            memset(cap, 0, sizeof(cap));            memset(cost, 0, sizeof(cost));            for (int i = 1; i <= M; i++) //源点向供应点建边                cap[start][i] = b[i][kk];            for(int i = 1; i <= N; i++) //店主向汇点建边                cap[M + i][End] = a[i][kk];            for (int i = 1; i <= N; i++)                for (int j = 1; j <= M; j++)                {                    scanf("%d", &tmp);                    cost[j][i + M] = tmp;                    cost[i + M][j] = -tmp;                    cap[j][i + M] = 1000000;                }            minCostMaxflow();            ans += c;            res += f;        }        if (res == need)            printf("%d\n", ans);        else            printf("-1\n");    }    return 0;}

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