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;

}