poj2516--Minimum Cost(费用流，分别建图)

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 13531		Accepted: 4635

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

给出的物品的种类太多，不同的供应商对不同的店家供应不同的货物的花费不同，这样是同一个图中的边太多，直接建图会超时，所以，对每一种货物分别建图，求出每一种的货物的最小花费，得到最终的花费

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define maxn 60
#define INF 0x3f3f3f3f
struct node
{
    int v , w , s ;
    int next ;
} p[100000];
int nk[maxn][maxn] , mk[maxn][maxn] , cnt ;//nk存储不同的店家对不同货物的需求，mk，不同的供应商对不同货物的供应量
int head[maxn<<2] , vis[maxn<<2] , dis[maxn<<2] , pre[maxn<<2] ;
queue <int> q ;
void add(int u,int v,int w,int s)
{
    p[cnt].v = v ;
    p[cnt].w = w ;
    p[cnt].s = s ;
    p[cnt].next = head[u] ;
    head[u] = cnt++ ;
    p[cnt].v = u ;
    p[cnt].w = 0 ;
    p[cnt].s = -s ;
    p[cnt].next = head[v] ;
    head[v] = cnt++ ;
}
int spfa(int s,int t)
{
    int u , v , i ;
    memset(dis,INF,sizeof(dis));
    vis[s] = 1 ; dis[s] = 0 ;
    pre[s] = pre[t] = -1 ;
    while( !q.empty() )
        q.pop();
    q.push(s);
    while( !q.empty() )
    {
        u = q.front();
        q.pop();
        vis[u] = 0 ;
        for(i = head[u] ; i != -1 ; i = p[i].next)
        {
            v=  p[i].v ;
            if( p[i].w && dis[v] > dis[u] + p[i].s )
            {
                dis[v] = dis[u] + p[i].s ;
                pre[v] = i ;
                if( !vis[v] )
                {
                    vis[v] = 1 ;
                    q.push(v);
                }
            }
        }
    }
    if( pre[t] == -1 )
        return 0;
    return 1 ;
}
int f(int s,int t,int k)
{
    int ans_s = 0 , ans_w = 0 , min1 , i ;//ans_s存储最小花费，ans_w存储一共的流量
    memset(vis,0,sizeof(vis));
    memset(pre,-1,sizeof(pre));
    while( spfa(s,t) )
    {
        min1 = INF ;
        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])
            if( p[i].w < min1 )
                min1 = p[i].w ;
        for(i = pre[t] ; i != -1 ; i = pre[ p[i^1].v ])
        {
            p[i].w -= min1 ;
            p[i^1].w += min1 ;
            ans_s += min1*p[i].s ;
        }
        ans_w += min1 ;
    }
    if(ans_w == nk[0][k])
        return ans_s ;
    return -1;
}
int main()
{
    int i , j , n , m , kk , k , x , ans , flag ;//输入很复杂，尤其是不同的供应商对不同的店家供应不同的货物有不同的价格。
    while(scanf("%d %d %d", &n, &m, &k) && n+m+k != 0)
    {
        /*建图方式，对每种货物分别建图，对于第ki种货物的建图：由源点到每个供应商（容量是供应的货物的量，花费是0），由供应商到店家（容量是INF，花费是供应ki货物的价格）
        ，有店家到汇点建图（容量是店家需要的货物的量，花费是0）*/
        memset(nk,0,sizeof(nk));
        for(i = 1 ; i <= n ; i++)
            for(j = 1 ; j <= k ; j++)
            {
                scanf("%d", &nk[i][j]);
                nk[0][j] += nk[i][j] ;//nk[0][]储存下一共需要第j样货物的数量
            }
        for(i = 1 ; i <= m ; i++)
            for(j = 1 ; j <= k ; j++)
                scanf("%d", &mk[i][j]);
        ans = flag = 0 ;//flag判断之前的货物是不是都可以供应足够，如果不够，不用向下在计算。
        for(kk = 1 ; kk <= k ; kk++)
        {
            cnt = 0 ;
            memset(head,-1,sizeof(head));
            for(i = 1 ; i <= m ; i++)
                add(0,i,mk[i][kk],0);
            for(i = 1 ; i <= n ; i++)
                add(m+i,m+n+1,nk[i][kk],0);
            for(i = 1 ; i <= n ; i++)
                for(j = 1 ; j <= m ; j++)
                {
                    scanf("%d", &x);
                    add(j,m+i,INF,x);
                }
            if(flag != -1)
            {
                flag = f(0,n+m+1,kk);
                if(flag != -1)
                    ans += flag ;
            }
        }
        if(flag == -1)
            printf("-1\n");
        else
            printf("%d\n", ans);
    }
    return 0;
}