POJ2516 Minimum Cost —— 最小费用最大流

题目链接：https://vjudge.net/problem/POJ-2516

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 17650		Accepted: 6205

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

题意：

有n个商店，m个仓库，k中商品。每个商店对每种商品都有特定需求量，且每个仓库中，每种商品都有其特定的存量。且已知对于某一种商品G，从仓库A运送一件商品G到商店B的运费。问：能否满足所有商店的供货需求？如果能满足，求出最小总运费？

题解：

最小费用最大流问题。可知每一种商品是相互独立的，因此我们可以单独求出每种商品总的最小运费，然后加起来，当然前提条件是能满足需求。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXM = 1e4+10;
const int MAXN = 1e2+10;

struct Edge
{
    int to, next, cap, flow, cost;
}edge[MAXM<<1];
int tot, head[MAXN];
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;

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

void add(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].next = head[u];   head[u] = tot++;
    edge[tot].to = u;   edge[tot].cap = 0;  edge[tot].cost = -cost;
    edge[tot].flow = 0; edge[tot].next = head[v];   head[v] = tot++;
}

bool spfa(int s, int t)
{
    queue<int>q;
    for(int i = 0; i<N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }

    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u  = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i!=-1; 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;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t]==-1) return false;
    return true;
}

int minCostMaxFlow(int s, int t, int &cost)
{
    int flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        int Min = INF;
        for(int i = pre[t]; i!=-1; i = pre[edge[i^1].to])
        {
            if(Min>edge[i].cap-edge[i].flow)
                Min = edge[i].cap-edge[i].flow;
        }
        for(int i = pre[t]; i!=-1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost*Min;
        }
        flow += Min;
    }
    return flow;
}

int need[55][55], storage[55][55], fee[55][55][55];
int main()
{
    int n, m, k;
    while(scanf("%d%d%d",&n,&m,&k)&&(n||m||k))
    {
        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", &storage[i][j]);

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

        int cost = 0;
        for(int item = 1; item<=k; item++)
        {
            int whole = 0;
            for(int i = 1; i<=n; i++)
                whole += need[i][item];

            int start = 0, end = n+m+1;
            init(n+m+2);
            for(int i = 1; i<=m; i++)
            {
                add(start, i, storage[i][item], 0);
                for(int j = 1; j<=n; j++)
//                 if(storage[i][item]>=need[j][item])  //不能加此条判断
                    add(i, m+j, storage[i][item], fee[item][j][i]);
            }

            for(int i = 1; i<=n; i++)
                add(m+i, end, need[i][item], 0);

            int cost_item;
            int offered = minCostMaxFlow(start, end, cost_item);
            if(offered<whole)
            {
                cost = -1;
                break;
            }
            else cost += cost_item;
        }
        printf("%d
", cost);
    }
}