POJ 2516--Minimum Cost【最小费用最大流 && 经典】

Minimum Cost

Time Limit: 4000MS Memory Limit: 65536KTotal Submissions: 15041 Accepted: 5169

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 10 1 11 2 21 0 11 2 31 1 12 1 11 1 132200 0 0

Sample Output

4-1

题意：

有N个店家、M个仓库以及K种物品。接下来N行每行K个数，表示每个店家对每一种物品的需求。后面跟着M行，表示每个仓库所存储的每种物品的数目。
最后给出K个N*M的矩阵。对于第k个矩阵，它的第i行第j列表示——第j个仓库运送第k种物品到第i个店家的费用。现在问你能否满足所有店家的需求，若可以输出最小的花费，反之输出-1。

解析：

把K种物品分开单独求解最小费用最大流。如果有一种物品不满足店家的要求的话，说明这种情况不可以。如果都满足话，把每种物品需要的最小费用加起来，就是总的最小花费。这样就转化成常规的最小费用问题了，建图也是常规的建图。

对每种物品，我们都用一下建图方式：

（1）源点到每个仓库建边，容量为拥有该种商品的数目，费用为0。

（2）仓库到店铺建图，容量为店铺对该种物品的需求，费用题目已给出。

（3）店铺到汇点建图，容量为对该种物品的需求，费用为0。

求出每一种物品的最小花费后，我们都要判断一下仓库中拥有该物品的数量能否满足店铺的需求。不满足的话，说明这种情况不可以。如果都满足话，把每种物品需要的最小费用加起来，就是总的最小花费。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define maxn 500#define maxm 11000#define INF 0x3f3f3f3fusing namespace std;int n, m, k;int inset, outset;int sum;struct node {    int u, v, cap, flow, cost, next;};node edge[maxm];int head[maxn], cnt;int dist[maxn], vis[maxn];int per[maxn];int need[100][100];int have[100][100];int map[100][100];int sumneed[100];void init(){    cnt = 0;    memset(head, -1, sizeof(head));}void add(int u, int v, int w, int c){    node E1 = {u, v, w, 0, c, head[u]};    edge[cnt] = E1;    head[u] = cnt++;    node E2 = {v, u, 0, 0, -c, head[v]};    edge[cnt] = E2;    head[v] = cnt++;}void input(){    memset(sumneed, 0, sizeof(sumneed));    for(int i = 1; i <= n; ++i)    for(int j = 1; j <= k; ++j){        scanf("%d", &need[i][j]);        sumneed[j] += need[i][j];    }    for(int i = 1; i <= m; ++i)    for(int j = 1; j <= k; ++j){        scanf("%d", &have[i][j]);    }}bool SPFA(int st, int ed){    queue<int>q;    for(int i = 0; i <= inset; ++i){        dist[i] = INF;        vis[i] = 0;        per[i] = -1;    }    dist[st] = 0;    vis[st] = 1;    q.push(st);    while(!q.empty()){        int u = q.front();        q.pop();        vis[u] = 0;        for(int i = head[u]; i != -1; i = edge[i].next){            node E = edge[i];            if(dist[E.v] > dist[u] + E.cost && E.cap > E.flow){                dist[E.v] = dist[u] + E.cost;                per[E.v] = i;                if(!vis[E.v]){                    vis[E.v] = 1;                    q.push(E.v);                }            }        }    }    return per[ed] != -1;}void MCMF(int st, int ed, int &cost, int &flow){    flow = 0;    cost = 0;    while(SPFA(st, ed)){        int mins = INF;        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){            mins = min(mins, edge[i].cap - edge[i].flow);        }        for(int i = per[ed]; i != -1; i = per[edge[i ^ 1].v]){            edge[i].flow += mins;            edge[i ^ 1].flow -= mins;            cost += edge[i].cost * mins;        }        flow += mins;    }}void solve(){    outset = 0;    inset = n + m + 1;    sum = 0;    int flag = 1;    for(int ii = 1; ii <= k; ++ii){        init();        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= m; ++j)                scanf("%d", &map[i][j]);        for(int i = 1; i <= m; ++i)            add(outset, i, have[i][ii], 0);//源点向仓库建边        for(int i = 1; i <= n; ++i)            add(i + m, inset, need[i][ii], 0);//店铺向汇定建边        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= m; ++j)            add(j, i + m, INF, map[i][j]);//仓库向店铺建边        int cost, flow;        MCMF(outset, inset, cost, flow);        if(flow == sumneed[ii])//注意            sum += cost;        else            flag = 0;    }    if(flag)        printf("%d\n", sum);    else        printf("-1\n");}int main (){    while(scanf("%d%d%d", &n, &m, &k), n || m || k){        input();        solve();    }    return 0;}