Poj 2516 Minimum Cost (最小花费最大流)

题目链接：

　　Poj  2516  Minimum Cost

题目描述：

　　有n个商店，m个仓储，每个商店和仓库都有k种货物。嘛！现在n个商店要开始向m个仓库发出订单了，订单信息为当前商店对每种货物的需求量。不同的商店从不同的仓库购买不同货物花费不同，问是不是能满足所有商店的要求，如果能商店花费总和最小为多少？

解题思路：

　　简单的费用流，要跑K次最小花费最大流，每次只对一种货物建图跑费用流。每次建图以0为源点, [1,m]为仓库, [m+1, n+m]为商店, n+m+1为汇点。0与[1,m]连边，边容量为仓库存货，边单位流量的花费为0。

[1,m]与[m+1,n+m]连边，边容量为INF，边单位流量的花费为对应仓库供应i货物给对应商店的单位花费。[m+1,n+m]与汇点连边，边容量为商店需求，边单位流量的花费为0。这个题目建图时候的初始化一定要搞好，不然很容易TLE。

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 210;
const int INF = 0x3f3f3f3f;
int flow[maxn][maxn], cap[maxn][maxn];
int shop[maxn][maxn], supply[maxn][maxn];
int vis[maxn], pre[maxn], dis[maxn], cost[maxn][maxn];
int c, Flow, s, e;

bool spfa ()
{
    queue <int> Q;
    for (int i=s; i<=e; i++)
    {
        dis[i] = INF;
        vis[i] = 0;
    }
    dis[0] = 0;
    vis[0] = 1;
    Q.push (s);
    while (!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        vis[u] = 0;
        for (int v=s; v<=e; v++)
        {
            if (cap[u][v]>flow[u][v] && dis[v]>dis[u]+cost[u][v])
            {
                dis[v] = dis[u] + cost[u][v];
                pre[v] = u;
                if (!vis[v])
                {
                    vis[v] = 1;
                    Q.push(v);
                }
            }
        }
    }
    if (dis[e] == INF)
        return false;
    return true;
}

void MincostMaxflow ()
{
    memset (flow, 0, sizeof(flow));
    c = Flow = 0;
    while (spfa())
    {
        int Min = INF;
        for (int i=e; i!=s; i=pre[i])
            Min = min (Min, cap[pre[i]][i]-flow[pre[i]][i]);
        for (int i=e; i!=s; i=pre[i])
        {
            flow[pre[i]][i] += Min;
            flow[i][pre[i]] -= Min;
        }
        Flow += Min;
        c += Min * dis[e];
    }
}

int main ()
{
    int n, m, k;
    while (scanf ("%d %d %d", &n, &m, &k), n||m||k)
    {
        int flag, Cost;
        flag = Cost = s = 0, e = n + m + 1;
        
        for (int i=1; i<=n; i++)
            for (int j=1; j<=k; j++)
                scanf ("%d", &shop[i][j]);

        for (int i=1; i<=m; i++)
            for (int j=1; j<=k; j++)
                scanf ("%d", &supply[i][j]);
                
        memset (cap, 0, sizeof(cap));
        memset (cost, 0, sizeof(cost));
        for (int i=1; i<=m; i++)
            for (int j=m+1; j<=n+m; j++)
                cap[i][j] = INF;

        for (int i=1; i<=k; i++)
        {
            int total = 0;
            for (int j=1; j<=n; j++)
                for (int l=1; l<=m; l++)
                {
                    scanf ("%d", &cost[l][j+m]);
                    cost[j+m][l] = -cost[l][j+m];
                }
                
            if (flag)
                continue;

            for (int j=1; j<=n; j++)
                {
                    cap[j+m][e] = shop[j][i];
                    total += shop[j][i];
                }
            for (int j=1; j<=m; j++)
                cap[s][j] = supply[j][i];

            MincostMaxflow();
            if (Flow < total)
                flag = 1;
            Cost += c;
            
        }
        printf ("%d
", flag?-1:Cost);
    }
    return 0;
}