POJ2516 Minimum Cost【最小费用最大流】

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;
const int maxn = 110;
const int inf = 99999999;

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

bool spfa()
{
    int i, head = 0, tail = 1;
    for(i = 0; i <= n; i ++){
        dis[i] = inf;
        vis[i] = false;
    }
    dis[0] = 0;
    que[0] = 0;
    while(tail != head){
        int u = que[head];
        vis[u] = true;
        for(i = 0; i <= n; i ++)
            if(cap[u][i] && dis[i] > dis[u] + cost[u][i]){
                dis[i] = dis[u] + cost[u][i];
                pre[i] = u;
                if(!vis[i]){
                    vis[i] = true;
                    que[tail ++] = i;
                    if(tail == maxn) tail = 0;
                }
            }
        vis[u] = false;
        head ++;
        if(head == maxn) head = 0;
    }
    if(dis[n] == inf) return false;
    return true;
}

void end()
{
    int i, sum = inf;
    for(i = n; i != 0; i = pre[i])
        sum = min(sum, cap[pre[i]][i]);
    for(i = n; i != 0; i = pre[i]){
        cap[pre[i]][i] -= sum;
        cap[i][pre[i]] += sum;
        ans += cost[pre[i]][i] * sum;
    }
}

int main()
{
    int N, M, K, i, j, k;
    int need[maxn][maxn], NeedK[maxn];
    int have[maxn][maxn], HaveK[maxn];
    while(cin>>N>>M>>K,N,M,K)
    {
        memset(NeedK, 0, sizeof(NeedK));
        for(i = 1; i <= N; i ++)
            for(j = 1; j <= K; j ++){
                scanf("%d", &need[i][j]);    //  第i个客户需要第j种货物的量。
                NeedK[j] += need[i][j];      //  第j种货物总共需要的量。
            }
        memset(HaveK, 0, sizeof(HaveK));
        for(i = 1; i <= M; i ++)
            for(j = 1; j <= K; j ++){
                scanf("%d", &have[i][j]);    //  第i个仓库存储第j种货物的量。
                HaveK[j] += have[i][j];      //  第j种货物总共需要的量。
            }
        bool flag = true;
        for(i = 1; i <= K; i ++)             //  判断所有货物是否足够。
            if(NeedK[i] > HaveK[i]){
                flag = false; break;
            }
        ans = 0;
        n = N + M + 1;
        for(k = 1; k <= K; k ++){
            memset(cap, 0, sizeof(cap));
            for(i = 1; i <= N; i ++)         //  建图。
                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;
            for(i = 1; i <= M; i ++){
                cap[0][i] = have[i][k];
                cost[0][i] = cost[i][0] = 0;
            }
            for(i = 1; i <= N; i ++){
                cap[M+i][n] = need[i][k];
                cost[M+i][n] = cost[n][M+i] = 0;
            }
            while(spfa()) end();            //  最小费用最大流算法。
        }
        if(flag) cout << ans << endl;
        else cout << -1 << endl;
    }
    return 0;
}