POJ2516:Minimum Cost(最小费用最大流)

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 19088		Accepted: 6740

题目链接：http://poj.org/problem?id=2516

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

题意：

这题理解好题意很关键，有n个商店，m个供应商，k种货物。

每个商店对每种货物都有一定的需求([0,3])，每个供应商也囤积了一定数量的每种货物([0,3])。

后面会输入k个矩阵，每个矩阵都是n*m的。

第K个矩阵里面，第i行第j列表示从j供应商到i商店运送第K种货物的花费。

最后求如若满足所有商店进货需求的最小花费，如果不能满足商店需求，就输出-1。

 

题解：

这里，每个供应商对于每种货物运送到不同的商店费用都是不一样的。

题目要求输入k个矩阵，我们考虑对于每种货物，都建一次图，源点连着所有的商店，供应点连着所有的汇点，边权都为他们所需要/拥有的货物个数，然后对于从j供应商到i商店运输的花费，都在其中间连一条容量为INF，费用为输入值的边。

最后跑k次最小费用流就行了。

另外注意的就是如果每次跑出来的最大流小于商店的需求，那么就是没解的。

最后，读懂题意很重要，我就是被坑在读题上面的...

 

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define t 200
#define INF 99999999
using namespace std;
typedef long long ll;
const int N = 205;
int n,m,k,tot;
int head[N],need[N][N],supply[N][N],d[N],a[N],vis[N],p[N],pre[N];
struct Edge{
    int v,next,c,w;
}e[(N*N)<<1];
void adde(int u,int v,int c,int w){
    e[tot].v=v;e[tot].next=head[u];e[tot].c=c;e[tot].w=w;head[u]=tot++;
    e[tot].v=u;e[tot].next=head[v];e[tot].c=0;e[tot].w=-w;head[v]=tot++;
}
int SPFA(int &flow,int &cost){
    for(int i=0;i<=t;i++) d[i]=a[i]=INF;d[0]=0;
    memset(vis,0,sizeof(vis));vis[0]=1;
    memset(p,-1,sizeof(p));memset(pre,-1,sizeof(pre));
    queue <int> q;q.push(0);
    while(!q.empty()){
        int u=q.front();q.pop();vis[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(e[i].c>0 && d[v]>d[u]+e[i].w){
                d[v]=d[u]+e[i].w;
                p[v]=u;pre[v]=i;
                a[v]=min(a[u],e[i].c);
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    if(d[t]==INF) return 0;
    flow+=a[t];cost+=d[t]*a[t];
    for(int i=t;i!=-1;i=p[i]){
        int edge=pre[i];
        e[edge].c-=a[t];
        e[edge^1].c+=a[t];
    }
    return 1;
}
int main(){
    while(scanf("%d%d%d",&n,&m,&k)!=EOF){
        if(!n &&!m &&!k) break;
        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 ans=0;
        bool flag=false;
        for(int K=1;K<=k;K++){
            tot=0;memset(head,-1,sizeof(head));
            for(int i=1;i<=n;i++){
                for(int j=1,s;j<=m;j++){
                    scanf("%d",&s);
                    adde(i,j+n,INF,s);
                }
            }
            int sum=0;
            for(int i=1;i<=n;i++) adde(0,i,need[i][K],0),sum+=need[i][K];
            for(int i=n+1;i<=n+m;i++) adde(i,t,supply[i-n][K],0);
            if(flag) continue ;
            int flow=0,cost=0;
            while(SPFA(flow,cost));
            if(sum>flow) flag=true;
            ans+=cost;
        }
        if(flag) puts("-1");
        else printf("%d\n",ans);
    }
    return 0;
}