POJ 2516 Minimum Cost 費用流

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 13052		Accepted: 4442

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

 

題目大意：n個顧客，m間商店，k件物品，每一個顧客對每件商品都有必定數量的需求，每間商店每一個物品也都有供應上限，且不一樣的商店對不一樣的顧客供應的不一樣的商品的價格不同（真繞T_T）。問知足全部顧客需求的最小花費，沒有就輸出-1。

題目分析：對每一件物品拆開來作最小費用流便可。若是滿流則有解，不然無解。PS：一開始不拆開來直接構圖TLE的停不下來Q_Q。。。

 

代碼以下：

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define REP(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
const int maxE = 3000000;
const int maxN = 300;
const int oo = 0x3f3f3f3f;
struct Edge{
    int v, c, w, n;
};
Edge edge[maxE];
int adj[maxN], l;
int d[maxN], cur[maxN], a[maxN];
int inq[maxN], Q[maxE], head, tail;
int n, m, k;
int cost, flow, s, t;
int nk[maxN][maxN], mk[maxN][maxN];
void addedge(int u, int v, int c, int w){
    edge[l].v = v; edge[l].c = c; edge[l].w =  w; edge[l].n = adj[u]; adj[u] = l++;
    edge[l].v = u; edge[l].c = 0; edge[l].w = -w; edge[l].n = adj[v]; adj[v] = l++;
}
int SPFA(){
    memset(d, oo, sizeof d);
    memset(inq, 0, sizeof inq);
    head = tail = 0;
    d[s] = 0;
    a[s] = oo;
    cur[s] = -1;
    Q[tail++] = s;
    while(head != tail){
        int u =  Q[head++];
        inq[u] = 0;
        for(int i = adj[u]; ~i; i = edge[i].n){
            int v = edge[i].v;
            if(!edge[i].c || d[v] <= d[u] + edge[i].w) continue;
            d[v] = d[u] + edge[i].w;
            cur[v] = i;
            a[v] = min(edge[i].c, a[u]);
            if(inq[v]) continue;
            inq[v] = 1;
            Q[tail++] = v;
        }
    }
    if(d[t] == oo) return 0;
    flow += a[t];
    cost += a[t] * d[t];
    for(int i = cur[t]; ~i; i = cur[edge[i ^ 1].v]){
        edge[i].c -= a[t];
        edge[i ^ 1].c += a[t];
    }
    return 1;
}
int MCMF(){
    while(SPFA());
    return flow;
}
void work(){
    int w, sum = 0, _sum = 0;
    REP(i, n) REP(j, k) scanf("%d", &nk[i][j]);
    REP(i, m) REP(j, k) scanf("%d", &mk[i][j]);
    REP(i, n) REP(j, k) sum += nk[i][j];
    REP(i, m) REP(j, k) _sum += mk[i][j];
    if(sum > _sum){
        REP(k1, k) REP(i, n) REP(j, m) scanf("%*d");
        printf("-1\n");
        return;
    }
    s = 0; t = m + n + 1;
    flow = cost = 0;
    REP(k1, k) {
        memset(adj, -1, sizeof adj);
        l = 0;
        REP(i, n) REP(j, m){
            scanf("%d", &w);
            addedge(i, j + n, oo, w);
        }
        REP(i, n) addedge(s, i, nk[i][k1], 0);
        REP(j, m) addedge(j + n, t, mk[j][k1], 0);
        MCMF();
    }
    printf("%d\n", flow == sum ? cost : -1);
}
int main(){
    while(~scanf("%d%d%d", &n, &m, &k) && (n || m || k)) work();
    return 0;
}

POJ 2516

 

題目分析：