POJ 2516 Minimum Cost（最小费用流）

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".

题目大意：N个客户M个仓库K种物品。已知每个客户需要的每种物品的数量，和每个仓库拥有的每种物品的数量，和每个仓库运送每种物品到每个顾客的花费，求满足所有顾客的最小花费。

思路：由于每个物品独立，分开每个物品建图。考虑物品x，建立附加源点S，从S到每个仓库连一条边，容量为该仓库拥有物品x的数量，费用为0；从每个客户连一条边到附加汇点T，容量为每个客户需要的物品x的数量，费用为0；从每个仓库连一条边到每个客户，容量为无穷大，费用为仓库到客户运输物品x的花费。求最小费用最大流，若都满流，K个物品相加就是答案。若有一个流不满，则输出-1（不能满足顾客需求）。

PS：记得读完数据。

PS2：再次提出稠密图应该用ZKW费用流。

PS3：ZKW费用流写挫了WA了一次，这玩意儿好难写……

代码（266MS）：

 #include <cstdio>

 #include <cstring>

 #include <queue>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 const int MAXV = ;

 const int MAXE = MAXV * MAXV;

 const int INF = 0x7fffffff;

 struct ZEK_FLOW {

     int head[MAXV], dis[MAXV];

     int next[MAXE], to[MAXE], cap[MAXE], cost[MAXE];

     int n, ecnt, st, ed;

     void init() {

         memset(head, , sizeof(head));

         ecnt = ;

     }

     void add_edge(int u, int v, int c, int w) {

         to[ecnt] = v; cap[ecnt] = c; cost[ecnt] = w; next[ecnt] = head[u]; head[u] = ecnt++;

         to[ecnt] = u; cap[ecnt] = ; cost[ecnt] = -w; next[ecnt] = head[v]; head[v] = ecnt++;

     }

     void SPFA() {

         for(int i = ; i <= n; ++i) dis[i] = INF;

         priority_queue<pair<int, int> > que;

         dis[st] = ; que.push(make_pair(, st));

         while(!que.empty()) {

             int u = que.top().second, d = -que.top().first; que.pop();

             if(d != dis[u]) continue;

             for(int p = head[u]; p; p = next[p]) {

                 int &v = to[p];

                 if(cap[p] && dis[v] > d + cost[p]) {

                     dis[v] = d + cost[p];

                     que.push(make_pair(-dis[v], v));

                 }

             }

         }

         int t = dis[ed];

         for(int i = ; i <= n; ++i) dis[i] = t - dis[i];

     }

     int minCost, maxFlow;

     bool vis[MAXV];

     int add_flow(int u, int aug) {

         if(u == ed) {

             maxFlow += aug;

             minCost += dis[st] * aug;

             return aug;

         }

         vis[u] = true;

         int now = aug;

         for(int p = head[u]; p; p = next[p]) {

             int &v = to[p];

             if(cap[p] && !vis[v] && dis[u] == dis[v] + cost[p]) {

                 int t = add_flow(v, min(now, cap[p]));

                 cap[p] -= t;

                 cap[p ^ ] += t;

                 now -= t;

                 if(!now) break;

             }

         }

         return aug - now;

     }

     bool modify_label() {

         int d = INF;

         for(int u = ; u <= n; ++u) if(vis[u]) {

             for(int p = head[u]; p; p = next[p]) {

                 int &v = to[p];

                 if(cap[p] && !vis[v]) d = min(d, dis[v] + cost[p] - dis[u]);

             }

         }

         if(d == INF) return false;

         for(int i = ; i <= n; ++i) if(vis[i]) dis[i] += d;

         return true;

     }

     int min_cost_flow(int ss, int tt, int nn) {

         st = ss, ed = tt, n = nn;

         minCost = maxFlow = ;

         SPFA();

         while(true) {

             while(true) {

                 for(int i = ; i <= n; ++i) vis[i] = ;

                 if(!add_flow(st, INF)) break;

             }

             if(!modify_label()) break;

         }

         return minCost;

     }

 } G;

 int n, m, k;

 int need[MAXV][MAXV], have[MAXV][MAXV], sum[MAXV];

 int mat[MAXV][MAXV];

 int main() {

     while(scanf("%d%d%d", &n, &m, &k) != EOF) {

         if(n ==  && m ==  && k == ) break;

         memset(sum, , sizeof(sum));

         for(int i = ; i <= n; ++i)

             for(int j = ; j <= k; ++j) scanf("%d", &need[i][j]), sum[j] += need[i][j];

         for(int i = ; i <= m; ++i)

             for(int j = ; j <= k; ++j) scanf("%d", &have[i][j]);

         int ans = ; bool flag = true;

         for(int x = ; x <= k; ++x) {

             for(int i = ; i <= n; ++i)

                 for(int j = ; j <= m; ++j) scanf("%d", &mat[i][j]);

             if(!flag) continue;

             G.init();

             int ss = n + m + , tt = ss + ;

             for(int i = ; i <= m; ++i) G.add_edge(ss, i, have[i][x], );

             for(int i = ; i <= n; ++i) G.add_edge(i + m, tt, need[i][x], );

             for(int i = ; i <= n; ++i)

                 for(int j = ; j <= m; ++j) G.add_edge(j, i + m, INF, mat[i][j]);

             ans += G.min_cost_flow(ss, tt, tt);

             flag = (G.maxFlow == sum[x]);

         }

         if(flag) printf("%d\n", ans);

         else puts("-1");

     }

 }

巴特西

POJ 2516 Minimum Cost（最小费用流）

最新文章

热门文章