Minimum Cost----最小费用流

最新推荐文章于 2023-04-04 10:56:57 发布

wk476855

最新推荐文章于 2023-04-04 10:56:57 发布

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文章标签： struct c

本文链接：https://blog.youkuaiyun.com/wk476855/article/details/7796771

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本文介绍了一种解决资源分配问题的最小费用最大流算法，并通过实例展示了如何应用该算法来优化资源分配过程，降低总成本。

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题目:http://poj.org/problem?id=2516

源代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define INF 1e9
#define MAXN 110

const int N = 200, M = 10010;

using namespace  std;

int n,m,k,ans;

int cost[MAXN][MAXN][MAXN];

int need[MAXN][MAXN],sale[MAXN][MAXN];

int sasum[MAXN],nesum[MAXN];

struct MinCostMaxFlow {
    int e, ev[M], nxt[M], head[N];
    int  cost[M], dist[N];
    int cap[M];
    int pnt[N], road[N], q[N], bg, ed;
    bool vis[N];

 void init(int n) {
   e = 0;
   for(int i=0;i<=n;i++)
    head[i]=-1;
  }

 void addedge(int u, int v, int f, int  c) { //u->v flow=f, cost=c
   ev[e]=v; cap[e]=f; cost[e]=c; nxt[e]=head[u]; head[u]=e++;
   ev[e]=u; cap[e]=0; cost[e]=-c; nxt[e]=head[v]; head[v]=e++;
   }

 bool spfa(int s, int t, int n) {
     for(int i=0;i<=n;i++)  {  dist[i]=INF;  vis[i]=0; }
     bg = ed = dist[s] = 0;
     pnt[s] = s; q[ed++] = s;
     while (bg != ed) {
         int u = q[bg++]; vis[u] = 0;
         if (bg == N) bg = 0;
         for (int i = head[u]; ~i; i = nxt[i]) {
            if (cap[i] <= 0)  continue;
            int v = ev[i];
            if (dist[v] > dist[u] + cost[i]) {
                dist[v] = dist[u] + cost[i];
                pnt[v] = u; road[v] = i;
            if (!vis[v]) {
                q[ed++] = v; vis[v] = 1;
                if(ed == N)ed = 0;
                }
            }
        }
    }
    return dist[t] != INF;
}

 void mincost(int s, int t, int n, int &f, int  &c) {
        c = f = 0;
        while(spfa(s, t, n)){
            int minf = INF;
            for(int u = t; u != s; u = pnt[u])
                minf = min(minf, cap[road[u]]);
                for(int u = t; u != s; u = pnt[u]){
                    cap[road[u]] -= minf;
                    cap[road[u] ^ 1] += minf;
                    }
                f += minf;
                c += minf * dist[t];
            }
    }
};

int solve(int x)
{
    MinCostMaxFlow ans;
    int ans1,ans2;

    ans.init(m+n+1);

    for(int i=1;i<=m;i++)
    ans.addedge(0,i,sale[i][x],0);

    for(int i=1;i<=m;i++)
      for(int j=1;j<=n;j++)
    ans.addedge(i,j+m,INF,cost[x][j][i]);

    for(int i=1;i<=n;i++)
    ans.addedge(m+i,m+n+1,need[i][x],0);

    ans.mincost(0,m+n+1,m+n+2,ans1,ans2);

    return ans2;
}

int  main()
{
   // freopen("D:\\a.txt","r",stdin);
  while(true)
  {
      memset(sasum,0,sizeof(sasum));
      memset(nesum,0,sizeof(nesum));

      scanf("%d %d %d",&n,&m,&k);
      if(!n&&!m&&!k)  break;
      for(int i=1;i<=n;i++)
      {
          for(int j=1;j<=k;j++)
          {
              scanf("%d",&need[i][j]);
              nesum[j]+=need[i][j];
          }
      }

      for(int i=1;i<=m;i++)
      {
          for(int j=1;j<=k;j++)
          {
              scanf("%d",&sale[i][j]);
              sasum[j]+=sale[i][j];
          }
      }

    for(int p=1;p<=k;p++)
      for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
         scanf("%d",&cost[p][i][j]);

    int flag=1;
    for(int i=1;i<=k;i++)
     if(sasum[i]<nesum[i])
     {
         printf("-1\n");
         flag=0;
        break;
     }
     if(!flag)  continue;
      ans=0;
      for(int i=1;i<=k;i++)
      {
          int tmp=solve(i);
             ans+=tmp;
      }
      if(flag)  printf("%d\n",ans);
  }
  return 0;
}