Time Limit: 4000MS | Memory Limit: 65536K | |
Total Submissions: 13052 | Accepted: 4442 |
Description
(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
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
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
#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; }