poj 2516Minimum Cost

http://poj.org/problem?id=2516

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<queue>
  5 #define maxn 110
  6 using namespace std;
  7 
  8 int need[maxn][maxn];
  9 int sup[maxn][maxn];
 10 int cost[maxn][maxn][maxn];
 11 int cap[maxn][maxn];
 12 int flow[maxn][maxn];
 13 int cost1[maxn][maxn];
 14 int d[maxn];
 15 int p[maxn];
 16 int n,m,k;
 17 const int inf=1<<30;
 18 
 19 int main()
 20 {
 21     while(scanf("%d%d%d",&n,&m,&k)!=EOF){
 22         memset(p,0,sizeof(p));
 23         memset(cost1,0,sizeof(cost1));
 24         memset(need,0,sizeof(need));
 25         if(n==0&&m==0&&k==0) break;
 26         for(int i=1; i<=n; i++){
 27             for(int j=1; j<=k; j++)
 28             {
 29                 scanf("%d",&need[i][j]);
 30             }
 31         }
 32         for(int i=1; i<=m; i++)
 33         {
 34             for(int j=1; j<=k; j++)
 35             {
 36                 scanf("%d",&sup[i][j]);
 37             }
 38         }
 39         for(int c=1; c<=k; c++)
 40         {
 41             for(int i=1; i<=n; i++)
 42             {
 43                 for(int j=1; j<=m; j++)
 44                 {
 45                     scanf("%d",&cost[c][j][i]);
 46                 }
 47             }
 48         }
 49         int s=0,t=n+m+1;
 50         bool flag=true;
 51         int tt=0;
 52         for(int i=1; i<=k; i++)
 53         {
 54             memset(cap,0,sizeof(cap));
 55             for(int j=1; j<=m; j++)
 56                  cap[s][j]=sup[j][i];
 57             for(int j=1; j<=m; j++)
 58             {
 59                 for(int c=1; c<=n; c++)
 60                 {
 61                     cap[j][m+c]=sup[j][i];
 62                 }
 63             }
 64             for(int j=1; j<=m; j++)
 65                cap[j+m][t]=need[j][i];
 66             memset(flow,0,sizeof(flow));
 67             for(int j=1; j<=m; j++)
 68             {
 69                 for(int c=1; c<=n; c++)
 70                 {
 71                     cost1[j][c+m]=cost[i][j][c];
 72                     cost1[c+m][j]=-cost1[j][c+m];
 73                 }
 74             }
 75             queue<int>q;
 76             memset(flow,0,sizeof(flow));
 77             int mm=0;
 78             for(;;)
 79             {
 80                 bool inq[maxn];
 81                 for(int i=0; i<=n+m+1; i++) d[i]=(i==s?0:inf);
 82                 memset(inq,false,sizeof(inq));
 83                 q.push(s);
 84                 while(!q.empty())
 85                 {
 86                     int u=q.front(); q.pop();
 87                     inq[u]=false;
 88                     for(int v=0; v<=n+m+1; v++) if(cap[u][v]>flow[u][v]&&d[v]>d[u]+cost1[u][v])
 89                     {
 90                         d[v]=d[u]+cost1[u][v];
 91                         p[v]=u;
 92                         if(!inq[v])
 93                         {
 94                             inq[v]=true;
 95                             q.push(v);
 96                         }
 97                     }
 98                 }
 99                 if(d[t]==inf) break;
100                 int a=inf;
101                 for(int u=t; u!=s; u=p[u])
102                 {
103                     if(cap[p[u]][u]-flow[p[u]][u]<a)
104                     {
105                         a=cap[p[u]][u]-flow[p[u]][u];
106                     }
107                 }
108                 for(int u=t; u!=s; u=p[u])
109                 {
110                     flow[p[u]][u]+=a;
111                     flow[u][p[u]]-=a;
112                 }
113                 mm+=d[t]*a;
114             }
115             for(int j=1; j<=n; j++)
116             {
117                 if(flow[j+m][t]!=cap[j+m][t])
118                 {
119                     flag=false;
120                     break;
121                 }
122             }
123             if(!flag) break;
124             if(flag) tt+=mm;
125         }
126         if(flag) printf("%d\n",tt);
127         else printf("-1\n");
128     }
129     return 0;
130 }
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posted @ 2013-11-28 17:19  null1019  阅读(170)  评论(0)  编辑  收藏  举报