poj-2516.minimum cost(k次费用流)

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 19883		Accepted: 7055

Description

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport. 

It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place. 

Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper. 

The input is terminated with three "0"s. This test case should not be processed.

Output

For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

Sample Input

1 3 3   
1 1 1
0 1 1
1 2 2
1 0 1
1 2 3
1 1 1
2 1 1

1 1 1
3
2
20

0 0 0

Sample Output

4
-1

Source

POJ Monthly--2005.07.31, Wang Yijie

都在代码里了，不不建议抄袭代码，代码里有些调试代码，有需要的可以看代码之前注释，前面是解释，精髓在最后三行。

/*
    本题心得：一开始做题就有种感觉需要对商品拆点，然后满足每个商人，但是这样的话每个商品要拆为n个点，必然会有很大的空间浪费造成tle，
    实在没思路之后看了博客，看到说每种商品都是独立的，意思就是把商人需要的每种物品都单独买，然后统计最后结果就行了，这里有一个细节就是，
    因为目的是满足所有商人情况下的最小费用，那也就是最小费用最大流，所以我们事先判断某种商品是否够用，如果够用，那么最大流一定是满载的，
    所以不必担心找不到最大流，就找最小花费就行了。
    对于每个商品，我们记得要清空head数组，额贼，这个把我坑了好久，后来想如果不清空必然会在spfa中造成无限循环(想想为什么？)，所以对每件
    商品都需要init,对于每件商品，我们建立超级源点指向那些供应商，容量为最大供应数目花费为0，对于每个商人，我们建立一条边指向超级汇点，容量为商人
    对这件商品的需求数目(限制每个商人得到的物品数)，花费为0，对于每个供应商和他的商人之间建立一条由供应商指向商人的边，cap为inf(由于前面我们已经限制了每个供应商可以提供的物品)
    花费为这个供应商对这个商人供应这件物品的cost，跑一波费用流就ojk了。
    这样我们就
        通过供应商 -> 商人 限制了价格 
        通过 s -> 供应商 限制了供应个数
        通过商人 -> t 限制了商人的需求数目。
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 50 + 5, maxm = 1e4 + 5, inf = 0x3f3f3f3f;
int want[maxn][maxn], supply[maxn][maxn], sumwant[maxn], sumsupply[maxn], costij[maxn][maxn][maxn];
struct Edge {
    int to, next, cap, flow, cost, from;
} edge[maxm];
int head[maxn << 1], tot;
int pre[maxn << 1], dis[maxn << 1];
bool vis[maxn << 1];

int N;

void init(int n) {
    N = n;
    tot = 0;
    memset(head, -1, sizeof head);
}

void addedge(int u, int v, int cap, int cost) {
    edge[tot].to = v; edge[tot].cap = cap; edge[tot].cost = cost; edge[tot].flow = 0; edge[tot].from = u;
    edge[tot].next = head[u]; head[u] = tot ++;
    edge[tot].to = u; edge[tot].cap = 0; edge[tot].cost = -cost; edge[tot].flow = 0; edge[tot].from = v;
    edge[tot].next = head[v]; head[v] = tot ++;
}

bool spfa(int s, int t) {
    queue <int> que;
    // memset(dis, inf, sizeof dis);
    // memset(vis, false, sizeof vis);
    // memset(pre, -1, sizeof pre);
    for(int i = 0; i <= N; i ++) {
        dis[i] = inf;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    que.push(s);
    while(!que.empty()) {
        // printf("in bfs");
        int u = que.front();
        que.pop();
        vis[u] = false;
        for(int i = head[u]; ~i; i = edge[i].next) {
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost) {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v]) {
                    vis[v] = true;
                    // printf("now in push of bfs");
                    que.push(v);
                }
            }
        }
    }
    return ~pre[t];
    // if(pre[t] == -1) return false;
    // else return true;
}

int mincostmaxflow(int s, int t) {
    int cost = 0;
    while(spfa(s, t)) {
        // printf("spfa is true");
        int Min = inf;
        for(int i = pre[t]; ~i; i = pre[edge[i ^ 1].to]) {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
            // printf("now is find min");
        }
        for(int i = pre[t]; ~i; i = pre[edge[i ^ 1].to]) {
            edge[i].flow += Min;
            edge[i ^ 1].flow -= Min;
            cost += edge[i].cost * Min;
            // printf("now is update");
        }
    }
    return cost;
}

int main() {
    int n, m, k;
    while(~scanf("%d %d %d", &n, &m, &k) && (n | m | k)) {

        memset(want, 0, sizeof want);
        memset(supply, 0, sizeof supply);
        memset(sumwant, 0, sizeof sumwant);
        memset(sumsupply, 0, sizeof sumsupply);
        for(int i = 1; i <= n; i ++) {
            for(int j = 1; j <= k; j ++) {
                scanf("%d", &want[i][j]);
                sumwant[j] += want[i][j];
            }
        }
        for(int i = 1; i <= m; i ++) {
            for(int j = 1; j <= k; j ++) {
                scanf("%d", &supply[i][j]);
                sumsupply[j] += supply[i][j];
            }
        }
        bool flag = true;
        for(int i = 1; i <= k; i ++) {
            if(sumwant[i] > sumsupply[i]) {
                flag = false;
                break;
            }
        }
        for(int q = 1; q <= k; q ++) {
            for(int i = 1; i <= n; i ++) {
                for(int j = 1; j <= m; j ++) {
                    scanf("%d", &costij[q][i][j]);//第q件物品，第i个人从第j个供应商的花费
                }
            }
        }
        int s = 0, t = m + n + 1, mcmf = 0;
        if(flag) {
            for(int q = 1; q <= k; q ++) {
                init(n + m + 2);
                // printf("***************
");
                for(int i = 1; i <= m; i ++) {
                    addedge(s, i, supply[i][q], 0);
                }
                for(int i = 1; i <= n; i ++) {
                    addedge(i + m, t, want[i][q], 0);
                }
                for(int i = 1; i <= n; i ++) {
                    for(int j = 1; j <= m; j ++) {
                        addedge(j, i + m, inf, costij[q][i][j]);
                    }
                }
                // for(int i = 0; i < tot; i ++) {
                //     printf("%d -> %d
", edge[i].from, edge[i].to);
                // }
                mcmf += mincostmaxflow(s, t);
            }
            printf("%d
", mcmf);
        } else printf("-1
");
    }
    return 0;
}