poj2516

网络流有点寸步难行了，http://blog.sina.com.cn/s/blog_6635898a0100pabw.html 看这个博客写的代码，自己大致理解了一下，作为最小费用流的邻接矩阵实现的模板。

//288K 250ms
#include <iostream>
using namespace std;

const int Max = 110;
const int inf = 0x7fffffff;

int cost[Max][Max];
int cap[Max][Max];
int que[Max], pre[Max];
bool vis[Max];
int dis[Max];
int n, ans;

bool spfa()
{
	int i, f, r;

	memset(vis, false, sizeof(vis));
	for(i = 0; i <= n; i++)
		dis[i] = inf;
	que[0] = 0; 
	dis[0] = 0;
	f = 0;
	r = 1;
	while(f != r)
	{
		int cur = que[f++];
		if(f == Max) f = 0;
		vis[cur] = false;
		for(i = 0; i <= n; i++)
			if(cap[cur][i] && dis[i] > dis[cur] + cost[cur][i])
			{
				dis[i] = dis[cur] + cost[cur][i];
				pre[i] = cur;
				if(!vis[i])
				{
                   vis[i] = true;
				   que[r++] = i;
				   if(r == Max) r = 0;
				}
			}
	}
	if(dis[n] == inf) return false;
	else return true;
}

void end()
{
	int i, tmp = inf;
    
	for(i = n; i; i = pre[i])
		if(tmp > cap[pre[i]][i])
			tmp = cap[pre[i]][i];
    for(i = n; i; i = pre[i])
	{
		cap[pre[i]][i] -= tmp;
		cap[i][pre[i]] += tmp;
		ans += tmp*cost[pre[i]][i];
	}
}
int main()
{
	int i, j, k;
	int N, M, K;
	int order[Max/2][Max/2], storage[Max/2][Max/2];
    int needK[Max/2], haveK[Max/2];
    bool flag;

	freopen("a.txt", "r", stdin);
    while(scanf("%d%d%d", &N, &M, &K) && N)
	{
		memset(needK, 0, sizeof(needK));
		memset(haveK, 0, sizeof(haveK));
		n = N+M+1;   //  超级汇点
		ans = 0;
        for(i = 1; i <= N; i++)
		{
			for(j = 1; j <= K; j++)
			{
			   scanf("%d", &order[i][j]);// 第i个客户需要第j种货物的量
               needK[j] += order[i][j];
			}
		}
		for(i = 1; i <= M; i++)
		{
			for(j = 1; j <= K; j++)
			{
				scanf("%d", &storage[i][j]);
				haveK[j] += storage[i][j];
			}
		}
		flag = true;
		for(i = 1; i <= K; i++)
	        if(needK[i] > haveK[i])
			{
				flag = false;
				break;
			}
		for(k = 1; k <= K; k++)// 枚举k种货物，对每种货物求最大流
		{
			memset(cap, 0, sizeof(cap));
			for(i = 1; i <= N; i++)
				for(j = 1; j <= M; j++)
				{
					scanf("%d", &cost[j][M+i]);
					cost[M+i][j] = -cost[j][M+i];
					cap[j][M+i] = inf;
				}
			if(!flag) continue; // 当货物不够时，返回继续输入数据
            for(i = 1; i <= M; i++)
			{
				cap[0][i] = storage[i][k];
				cost[0][i] = cost[i][0] = 0;// 从超级源点到每一个货物提供点的运费为0
			}
			for(i = 1; i <= N; i++)
			{
				cap[i+M][n] = order[i][k];
				cost[i+M][n] = cost[n][i+M] = 0;
			}
			while(spfa()) {end();}
		}
		if(flag) printf("%d\n", ans);
		else  printf("-1\n");
	}
	return 0;
}