http://poj.org/problem?id=2516
题意:
有N的商店,M个仓库,k件商品,每个商店对于商品都有不同的数量需求,而仓库也存有不同的商品,每个仓库运送每个商品都有一个花费,问满足每个商家不最小费是多少
思路:
数据条件有点多,可以对每一种商品都进行一次最大流最小费,把每次的花费加起来,就可解了。
注意如果仓库的商品数量不够商家的需求,那么输出为-1,特判一下即可
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <stdio.h>
#include <deque>
using namespace std;
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define maxn 100005
#define eps 0.00000001
#define PI acos(-1.0)
#define M 1000000007
struct Edge{
int v, w, cap, cost, nxt;
}edge[maxn];
int tot, pre[maxn], head[maxn], vis[maxn], dis[maxn], S, T;
int n, m, k;
int Want[60][60], Sup[60][60], Cost[60][60][60];
void init() {
tot = 0;
memset(head, -1, sizeof(head));
S = 0; T = n + m + 1 ;
}
void addEdge(int u, int v, int w, int cost) {
edge[tot].v = v;
edge[tot].w = w;
edge[tot].cap = 0;
edge[tot].cost = cost;
edge[tot].nxt = head[u];
head[u] = tot ++;
edge[tot].v = u;
edge[tot].w = 0;
edge[tot].cost = -cost;
edge[tot].cap = 0;
edge[tot].nxt = head[v];
head[v] = tot ++;
}
bool SPFA() {
memset(dis, INF, sizeof(dis));
memset(vis, 0, sizeof(vis));
memset(pre, -1, sizeof(pre));
queue<int> que;
que.push(S);
dis[S] = 0;
vis[S] = 1;
pre[S] = -1;
while(!que.empty()) {
int u = que.front(); que.pop();
vis[u] = 0;
for (int i = head[u]; i + 1; i = edge[i].nxt) {
if(dis[edge[i].v] > dis[u] + edge[i].cost && edge[i].w > edge[i].cap) {
dis[edge[i].v] = dis[u] + edge[i].cost;
pre[edge[i].v] = i;
if(!vis[edge[i].v]) {
vis[edge[i].v] = 1;
que.push(edge[i].v);
}
}
}
}
if(pre[T] == -1)return 0;
return 1;
}
int MinCostMaxFlow(int &cost) {
int Flow = 0, MinCost = INF;
cost = 0;
while(SPFA()) {
for (int i = pre[T]; i + 1; i = pre[edge[i ^ 1].v])
MinCost = min(MinCost, edge[i].w - edge[i].cap);
for (int i = pre[T]; i + 1; i = pre[edge[i ^ 1].v]) {
edge[i].cap += MinCost;
edge[i ^ 1].cap -= MinCost;
cost += edge[i].cost * MinCost;
}
Flow += MinCost;
}
return Flow;
}
int main(int argc, const char * argv[]) {
while (scanf("%d %d %d", &n, &m, &k) && (n || k || m)) {
int SumWant[maxn], SumSup[maxn];
memset(SumSup, 0, sizeof(SumSup));
memset(SumWant, 0, sizeof(SumWant));
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= k; j ++) {
scanf("%d", &Want[i][j]); //要货
SumWant[j] += Want[i][j];
}
for (int i = 1; i <= m; i ++)
for (int j = 1; j <= k; j ++) {
scanf("%d", &Sup[i][j]); //供货
SumSup[j] += Sup[i][j];
}
for (int i = 1; i <= k; i ++)
for (int j = 1; j <= n; j ++)
for (int z = 1; z <= m; z ++)
scanf("%d", &Cost[z][j][i]);
bool flag = 0;
for (int i = 1; i <= k; i ++)
if(SumSup[i] < SumWant[i]) {
flag = 1;
break;
}
if(!flag) {
int MinCost = 0;
for (int i = 1; i <= k; i ++) {
init();
for (int j = 1; j <= m; j ++)
addEdge(S, j, Sup[j][i], 0);
for (int j = 1; j <= n; j ++)
addEdge(j + m, T, Want[j][i], 0);
for (int j = 1; j <= m; j ++)
for (int z = 1; z <= n; z ++)
addEdge(j, z + m, Want[z][i], Cost[j][z][i]);
int cost;
MinCostMaxFlow(cost);
//printf("%d\n", cost);
MinCost += cost;
}
printf("%d\n", MinCost);
}else {
printf("-1\n");
}
}
return 0;
}