POJ 2516 Minimum Cost 最小费用流 难度:1

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 13511		Accepted: 4628

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

这一题的边有两个性质,

1 容量:供给-种类/需求-种类

2 价格: 供给-种类-需求

怎么想都想不到怎么保留这两种性质建边

看到小you的博客,可以分种类建图,恍然大悟

于是对每个种类分成两种边

1 源点->供给 需求->最终汇点 用于控制流量,价格为0

2 供给->需求 用于控制价格,流量为inf,

值得一提的是供给->需求是单向边,返回的边价格应该是负数,在这里卡了一次,还有每次流量应该是需求量

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
using namespace std;
const int maxn=;
const int maxm=;
const int  maxk=;
const int maxnum=;
const int inf =0x7fffffff;

int f[maxnum][maxnum];//s 151 t 152
int cons[maxk][maxnum];
int cost[maxk][maxnum][maxnum];
int e[maxnum][maxnum];
int len[maxnum];

const int sups=,supt=;
int sum,n,m,k;

int s1[maxk],s2[maxk];//s1 0-n-1 s2 n-n+m
bool input(){
    memset(cons,,sizeof(cons));
    memset(s1,,sizeof(s1));
    memset(s2,,sizeof(s2));

    sum=;
    if(scanf("%d%d%d",&n,&m,&k)!=)return false;
    if(n==&&m==&&k==)return false;
    for(int i=;i<n;i++){
        int c;
        for(int j=;j<k;j++){
            scanf("%d",&c);
            cons[j][i]=c;
            s1[j]+=c;
        }
    }
    for(int i=n;i<n+m;i++){
        int c;
        for(int j=;j<k;j++){
            scanf("%d",&c);
            cons[j][i]=c;
            s2[j]+=c;
        }
    }

    for(int i=;i<k;i++){
        for(int j=;j<n;j++){
            for(int ii=n;ii<n+m;ii++){
                int c;
                scanf("%d",&c);
                cost[i][j][ii]=c;cost[i][ii][j]=-c;
            }
        }
    }

    for(int i=;i<n;i++){
        for(int j=n;j<n+m;j++){
            e[i][j-n]=j;
            e[j][i]=i;
            e[i][m]=;
            e[j][n]=;
            e[][i]=i;
            e[][j-n]=j;
        }
    }
    fill(len,len+n,m+);
    fill(len+n,len+n+m,n+);
    len[]=n;
    len[]=m;
    return true;
}
void build(int kind){
    memset(f,,sizeof(f));
    for(int i=;i<n;i++)f[][i]=cons[kind][i];
    for(int i=n;i<n+m;i++)f[i][]=cons[kind][i];
    for(int i=;i<n;i++){
        for(int j=n;j<n+m;j++){
            f[i][j]=inf;
        }
    }
}
int d[maxnum],pre[maxnum];
bool vis[maxnum];
queue<int >que;
int mincostmaxflow(int s,int flow,int kind){
    build(kind);
    int res=;
    while(flow>){
        fill(d,d+,inf);
        memset(vis,,sizeof(vis));
        d[s]=;
        que.push(s);
        while(!que.empty()){
            int fr=que.front();que.pop();
            vis[fr]=false;
            for(int i=;i<len[fr];i++){
                int t=e[fr][i];
                if(f[fr][t]>&&d[t]>d[fr]+cost[kind][fr][t]){
                    d[t]=d[fr]+cost[kind][fr][t];
                    pre[t]=fr;
                    if(!vis[t]){
                        que.push(t);
                        vis[t]=true;
                    }
                }
            }
        }
        if(d[supt]==inf)return -;
        int sub=flow;
        for(int v=supt;v!=sups;v=pre[v]){
            sub=min(sub,f[pre[v]][v]);
        }
        flow-=sub;
        res+=sub*d[supt];
        for(int v=supt;v!=sups;v=pre[v]){
            f[v][pre[v]]+=sub;
            f[pre[v]][v]-=sub;
        }
    }
    return res;
}
int main(){
    while(input()){
        int ans=;
        bool sign=false;
        for(int i=;i<k;i++){
            if(s1[i]>s2[i]){printf("-1\n");sign=true;break;}
        }
        if(sign)continue;
        for(int i=;i<k;i++){
            int res=mincostmaxflow(,s1[i],i);
            if(res==-){printf("-1\n");sign=true;break;}
            ans+=res;
        }
        if(sign)continue;
        printf("%d\n",ans);

    }
    return ;
}