POJ - 2516 Minimum Cost （MCMF）

题目大意：有n间商店，m个货源，k种商品
现在给出每间商店所需的商品个数和每个货源能提供的商品数量，和货源运输货物到商店的代价
问能否满足所有商店的需求，且代价最小

解题思路：每种商品的运输是是不相关的，所以可以将每种商品的运输分离出来求
超级源点—货源，容量为供货量，代价为0
货源—商店,容量为INF，代价为运输代价
商店—超级汇点，容量为需求量，代价为0

每次进行判断，如果不能满足的话，直接赋值为-1，也就是说后面可以不用再跑了，因为不可能满足了，判断条件就是取反，-1取反刚好是0

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge{
    int from, to, cap, flow ,cost;
    Edge() {}
    Edge(int from, int to, int cap, int flow, int cost):from(from), to(to), cap(cap), flow(flow), cost(cost) {}
};

struct MCMF{
    int n, m, source, sink;
    vector<Edge> edges;
    vector<int> G[N];
    int d[N], f[N], p[N];
    bool vis[N];

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++)
            G[i].clear();
        edges.clear();
    }

    void AddEdge(int from, int to, int cap, int cost) {
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool BellmanFord(int s, int t, int &flow, int &cost) {
        for (int i = 0; i <= n; i++)
            d[i] = INF;
        memset(vis, 0, sizeof(vis));
        vis[s] = 1; d[s] = 0; f[s] = INF; p[s] = 0;
        queue<int> Q;
        Q.push(s);

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            vis[u] = 0;

            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    f[e.to] = min(f[u], e.cap - e.flow);
                    if (!vis[e.to]) {
                        vis[e.to] = true;
                        Q.push(e.to);
                    }
                }
            }
        }

        if (d[t] == INF)
            return false;

        flow += f[t];
        cost += d[t] * f[t];

        int u = t;
        while (u != s) {
            edges[p[u]].flow += f[t];
            edges[p[u] ^ 1].flow -= f[t];
            u = edges[p[u]].from;
        }
        return true;
    }

    int Mincost(int s, int t, int &cost) {
        int flow = 0;
        while (BellmanFord(s, t, flow, cost));
        return flow;
    }
};
MCMF mcmf;
#define M 60
int need[M][M], supply[M][M];
int n, m, k;

void init() {
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= k; j++)
            scanf("%d", &need[i][j]);

    for (int i = 1; i <= m; i++)
        for (int j = 1; j <= k; j++)
            scanf("%d", &supply[i][j]);

    int s = n + m + 1, t = n + m + 2;
    int cost = 0, tmp;

    for (int i = 1; i <= k; i++) {
        mcmf.init(t);
        int Sum = 0;

        for (int j = 1; j <= n; j++) {
            mcmf.AddEdge(j, t, need[j][i], 0);
            Sum += need[j][i];
        }

        for (int j = 1; j <= m; j++) 
            mcmf.AddEdge(s, j + n, supply[j][i], 0);
        for (int j = 1; j <= n; j++)
            for (int l = 1; l <= m; l++) {
                scanf("%d", &tmp);
                mcmf.AddEdge(n + l, j, INF, tmp);
            }
        if (~cost && mcmf.Mincost(s, t, cost) < Sum)
            cost = -1;
    }
    printf("%d\n", cost);
}

int main() {
    while (scanf("%d%d%d", &n, &m, &k) && n + m + k) {
        init();
    }
    return 0;
}