pku 2561 Minimum Cost 最小费用最大流

http://poj.org/problem?id=2516

刚接触，所以这道题目整了一天的时间，囧！！ 

题意描述也很难理解：n个店主，要从Dearboy的m个仓库里进k中商品。首先给出n个店主对k种的商品的需求量，然后给出Dearboy的m个仓库中分别存放k中货物的数量。最后给出的是从m个仓库输送k种商品到n个店主的费用。

由以上约束条件建立约束图，利用mcmf算法求解。。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define maxn 107
using namespace std;

const int inf = 99999999;

int sto[maxn][maxn],need[maxn][maxn],cost[maxn][maxn][maxn];
int pre[maxn],f[maxn][maxn],c[maxn][maxn],w[maxn][maxn];
int dis[maxn];
bool inq[maxn];
int n,m,k,s,t;

void input()
{
    int i,j,l;
    //输入n个店主的需求
    for (i = 1; i <= n; ++i)
    for (j = 1; j <= k; ++j) scanf("%d",&need[i][j]);
    //输入m个仓库存储的k中商品的数量
    for (i = 1; i <= m; ++i)
    for (j = 1; j <= k; ++j) scanf("%d",&sto[i][j]);
    //输入运输的花费
    for (l = 1; l <= k; ++l)
    for (i = 1; i <= n; ++i)
    for (j = 1; j <= m; ++j) scanf("%d",&cost[l][i][j]);
}
void init()
{
    memset(f,0,sizeof(f));
    memset(c,0,sizeof(c));
}
void spfa(int s)
{
    int v;
    queue<int>q;
    dis[s] = 0;
    q.push(s); inq[s] = true;
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        inq[u] = false;
        for (v = 0; v <= t; ++v)
        {
            if (c[u][v] > f[u][v] && dis[v] > dis[u] + w[u][v])
            {
                dis[v] = dis[u] + w[u][v];
                pre[v] = u;
                if (!inq[v])
                {
                    inq[v] = true;
                    q.push(v);
                }
            }
        }
    }
}
void mcmf(int s)
{
    int i;
    while (1)
    {
        for (i = 0; i < maxn; ++i)
        {
            dis[i] = inf; inq[i] = false;
            pre[i] = -1;
        }
        spfa(s);
        if (pre[t] == -1) break;
        int x = t,minf = inf;
        while (pre[x] != -1)
        {
            minf = min(minf,c[pre[x]][x] - f[pre[x]][x]);
            x = pre[x];
        }
        x = t;
        while (pre[x] != -1)
        {
            f[pre[x]][x] += minf;
            f[x][pre[x]] += minf;
            x = pre[x];
        }
    }
}
int main()
{
    int i,j,l;
    while (~scanf("%d%d%d",&n,&m,&k))
    {
        if (!n && !m && !k) break;
        input();
        s = 0; t = n + m + 1;
        bool flag = false;
        int ans = 0;
        //对每一种商品求最小费用最大流
        for (l = 1; l <= k && !flag; ++l)
        {
            //建图
            init();

            for (i = 1; i <= m; ++i) c[s][i] = sto[i][l];
            for (i = 1; i <= n; ++i) c[i + m][t] = need[i][l];

            for (i = 1; i <= m; ++i)
            for (j = 1; j <= n; ++j) c[i][j + m] = sto[i][l];

            for (i = 1; i <= m; ++i)
            {
                for (j = 1; j <= n; ++j)
                {
                    w[i][j + m] = cost[l][j][i];
                    w[j + m][i] = -w[i][j + m];
                }
            }
            mcmf(s);
            //判断是否可行
            for (i = 1; i <= n; ++i)
            {
                if (c[i + m][t] != f[i + m][t])
                {
                    flag = true; break;
                }
            }
            for (i = 1; i <= m; ++i)
            {
                for (j = 1; j <= n; ++j)
                {
                    ans += w[i][j + m]*f[i][j + m];
                }
            }
        }
        if (flag) puts("-1");
        else printf("%d\n",ans);
    }
    return 0;
}