poj 2516 Minimum Cost 【最小费用最大流】【求解K种物品的最小费用，独立求解累加每个结果】

Minimum Cost

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 15010		Accepted: 5154

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

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。

╮(╯▽╰)╭，一开始太任性了，把K种物品一起建边，然后跑一次MCMF求解最后答案，结果TLE5次。
最后把K种物品分开求解就AC了，跑了325ms。。。
   
   
   
   


   
   
   
   
思路：对每一种物品单独求解，累加结果即可。因为运输一种物品的最小费用是独立的，并且只要有一种物品不能满足所有店家的需求，就可以说明不存在一种可行方案。

建图：设置超级源点source，超级汇点sink。 这里只说明对第k种物品建边。
1，source到所有仓库建边，容量为该仓库所存储的第k种物品的数目，费用为0；
2，所有店家到sink建边，容量为店家对第k种物品的需求；
3，所有仓库到店家建边，容量为INF，因为我们可以任意选择通过该边的流量。至于费用（题目已给出）；
声明，所有店家对第k种物品的需求就是超级源点传入的总流量，需要记录下这个值来判断是否满流。
最后跑一次MCMF，看是否满流即满足所有店家的需求。若可以累加结果，反之说明不可能实现（只要有一种物品不满足所有店家的需求，就失败）。
当然为了节约时间，我们就可以记录对每个物品的总需求和总存储量，若有一个物品的总需求大于总存储量，就可以说明不存在可行方案。

AC代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 200
#define MAXM 10000+10
#define INF 0x3f3f3f3f
using namespace std;
struct Edge
{
    int from, to, cap, flow, cost, next;
};
Edge edge[MAXM];
int head[MAXN], edgenum;
int dist[MAXN], pre[MAXN];
bool vis[MAXN];
int source, sink;//超级源点 超级汇点
int N, M, K;//N个店主 M个仓库 K种食品
int need[60][60];//need[i][j] 第i个店主对第j种物品的需求
int have[60][60];//have[i][j] 第i个仓库里第j种物品的存货
int Sneed[60];//Sneed[i] 第i种物品的总需求
int Shave[60];//Shave[i] 第i种物品的总存货
int used[60][60];//在第k矩阵里面 存储i仓库到j店主的花费
bool flag;//判断能否满足所有人需求
void init()
{
    edgenum = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w, int c)
{
    Edge E1 = {u, v, w, 0, c, head[u]};
    edge[edgenum] = E1;
    head[u] = edgenum++;
    Edge E2 = {v, u, 0, 0, -c, head[v]};
    edge[edgenum] = E2;
    head[v] = edgenum++;
}
bool SPFA(int s, int t)
{
    queue<int> Q;
    memset(dist, INF, sizeof(dist));
    memset(vis, false, sizeof(vis));
    memset(pre, -1, sizeof(pre));
    dist[s] = 0;
    vis[s] = true;
    Q.push(s);
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            Edge E = edge[i];
            if(dist[E.to] > dist[u] + E.cost && E.cap > E.flow)
            {
                dist[E.to] = dist[u] + E.cost;
                pre[E.to] = i;
                if(!vis[E.to])
                {
                    vis[E.to] = true;
                    Q.push(E.to);
                }
            }
        }
    }
    return pre[t] != -1;
}
void MCMF(int s, int t, int &cost, int &flow)
{
    cost = flow = 0;
    while(SPFA(s, t))
    {
        int Min = INF;
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            Edge E = edge[i];
            Min = min(Min, E.cap-E.flow);
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost * Min;
        }
        flow += Min;
    }
}
void solve()
{
    memset(Sneed, 0, sizeof(Sneed));
    memset(Shave, 0, sizeof(Shave));
    for(int i = 1; i <= N; i++)//店主需求
    {
        for(int j = 1; j <= K; j++)//对每个物品的需求
        {
            scanf("%d", &need[i][j]);
            Sneed[j] += need[i][j];
        }
    }
    for(int i = 1; i <= M; i++)//每个仓库
    {
        for(int j = 1; j <= K; j++)//每个物品 存货
        {
            scanf("%d", &have[i][j]);
            Shave[j] += have[i][j];
        }
    }
    flag = true;
    for(int i = 1; i <= K; i++)
    {
        if(Shave[i] < Sneed[i])//存货 少于需求
        {
            flag = false;
            break;
        }
    }
    int ans = 0;//最后结果
    int cost, flow;
    for(int k = 1; k <= K; k++)//K个矩阵  对K种物品依次求解 因为每个解都是独立的
    {
        for(int i = 1; i <= N; i++)
        {
            for(int j = 1; j <= M; j++)
                scanf("%d", &used[i][j]);//第j个仓库运送第k种物品给第i个店主所需费用
        }
        if(!flag) continue;//已经判断出 不能满足需求
        init();
        source = 0, sink = N+M+1;
        for(int i = 1; i <= N; i++)//店主 向超级汇点建边
            addEdge(i, sink, need[i][k], 0);
        for(int i = 1; i <= M; i++)//超级源点 向仓库建边
            addEdge(source, i+N, have[i][k], 0);
        for(int i = 1; i <= M; i++)
        {
            for(int j = 1; j <= N; j++)
                addEdge(i+N, j, INF, used[j][i]);//仓库 向 每个店主建边
        }
        MCMF(source, sink, cost, flow);
        if(flow != Sneed[k])//总流量不等于 第k种物品的总需求
            flag = false;//不满足
        else
            ans += cost;//累加 每次的结果
    }
    if(flag)
        printf("%d\n", ans);
    else
        printf("-1\n");
}
int main()
{
    while(scanf("%d%d%d", &N, &M, &K), N||M||K)
    {
        solve();
    }
    return 0;
}