POJ-2516 Minimum Cost

题目链接：Minimum Cost

题目大意：有k种流通的物品，有n家店，m个供货商，每家店有k种物品的需求量，每个供货商也有k种物品的供应量，还有k个矩阵，第i行j列表示第j个供应商给第i个店第k种物品的费用。

解题思路：此题的关键思路在于将k个物品分开，每次选取一个物品做一次费用流。源点与供应商相连，容量为供应量，费用为0；汇点与店家相连，容量为需求量，费用为0；供应商与店家根据费用矩阵相连，容量为无穷大。跑k次费用流就可以得到总的最小费用了，因为不同物品的运输不会对彼此造成影响。注意在一开始时如果供应量小于需求量的话是无解的，我们可以做个标记，在连边的时候直接continue掉。

代码如下：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int inf  = 0x3f3f3f3f;
const int maxn = 2e3 + 15;

int need[55][55], supply[55][55], cost[55][55];

struct Edge {
    int to, cap, next, cost;
};

int ecnt, st, ed, ans;
Edge es[30000];
int head[500], dis[500], inq[500], pre[500], flow[500]; 

void init(){
    memset(head, -1, sizeof head);
    ecnt = 0;
}

void add_edge(int u, int v, int w, int cap) {
    es[ecnt].to = v;
    es[ecnt].next = head[u];
    es[ecnt].cap = cap;
    es[ecnt].cost = w;
    head[u] = ecnt++;
}

void add_double(int u, int v, int w, int cap) {
    add_edge(u, v, w, cap);
    add_edge(v, u, -w, 0);
}

int MCMF() {
    while(1) {
        memset(dis, 0x3f, sizeof dis);
        queue<int> que; que.push(st);
        inq[st] = 1; dis[st] = 0;
        flow[st] = inf;
        while(que.size()) {
            int u = que.front(); 
            que.pop(); inq[u] = 0;  
            for(int i = head[u]; ~i; i = es[i].next) {
                Edge& e = es[i]; 
                int v = e.to;
                if(e.cap && dis[v] > dis[u] + e.cost) {
                    dis[v] = dis[u] + e.cost;
                    flow[v] = min(flow[u], e.cap);
                    pre[v] = i ^ 1; 
                    if(inq[v]) continue;
                    inq[v] = 1; que.push(v);
                }
            }
        }
        if(dis[ed] == inf) return ans;
        ans += flow[ed] * dis[ed]; 
        for(int u = ed; u != st; u = es[pre[u]].to) {
            es[pre[u]].cap += flow[ed];
            es[pre[u] ^ 1].cap -= flow[ed];
        }
    }
    return -1;
}

inline int read() {
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c)) {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

int main(){
    int n, m, k;
    while(scanf("%d%d%d", &n, &m, &k) != EOF) {
        if(!n && !m && !k) break; 
        st = 0, ed = n + m + 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= k; j++)
                need[i][j] = read(); 

        for(int i = 1; i <= m; i++)
            for(int j = 1; j <= k; j++)
                supply[i][j] = read();

        bool fg = true; 
        for(int j = 1; j <= k; j++) {
            int l = 0, r = 0;
            for(int i = 1; i <= n; i++)
                l += need[i][j];    
            for(int i = 1; i <= m; i++)
                r += supply[i][j];  
            if(l > r) {
                fg = false;
                break;
            }
        }
        int sum = 0;
        for(int t = 1; t <= k; t++) {
            init(); ans = 0;
            for(int i = 1; i <= n; i++) {
                add_double(i + m, ed, 0, need[i][t]);   
                for(int j = 1; j <= m; j++) {
                    cost[j][i] = read();        
                    add_double(j, i + m, cost[j][i], inf);
                }
            }
            if(!fg) continue;
            for(int i = 1; i <= m; i++)
                add_double(st, i, 0, supply[i][t]);
            MCMF();                 
            sum += ans;             
        }
        if(!fg) puts("-1");
        else printf("%d\n", sum);
    }
    return 0;
}