HDU 4292 Food 多源多汇入门题

　　　　　　　　　　　　　　　　　　　　Food

有F种食物和D种饮料，每种食物或饮料只能供有限次，且每个人只享用一种食物和一种饮料。现在有n个人，每个人都有自己喜欢的食物种类列表和饮料种类列表，问最多能使几个人同时享用到自己喜欢的食物和饮料。

邻接矩阵 DINIC    在定点数较多的时候比较慢。time  608 ms   。邻接表可以加速

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <string>
  7 #include <vector>
  8 #include <set>
  9 #include <map>
 10 #include <stack>
 11 #include <queue>
 12 #include <sstream>
 13 #include <iomanip>
 14 using namespace std;
 15 typedef long long LL;
 16 const int INF = 0x4fffffff;
 17 const double EXP = 1e-5;
 18 const int MS = 205;
 19 const int SIZE = 100005;
 20
 21 int edges[4*MS][4*MS];
 22 int level[4*MS];
 23 int que[6*MS];
 24 char str[MS];
 25 int qs,qe;
 26 int n,f,d,cnt;
 27
 28 bool BFS()
 29 {
 30       memset(level,0xff,sizeof(level));
 31       level[0]=0;
 32       qs=qe=0;
 33       que[qe++]=0;
 34       while(qs<qe)
 35       {
 36             int u=que[qs++];
 37             for(int v=0;v<=cnt;v++)
 38             {
 39                   if(level[v]<0&&edges[u][v]>0)
 40                   {
 41                         level[v]=level[u]+1;
 42                         que[qe++]=v;
 43                   }
 44             }
 45       }
 46       return level[cnt]>0;
 47 }
 48
 49 int DFS(int u,int minv)
 50 {
 51       if(u==cnt)
 52             return minv;
 53       int t;
 54       for(int v=0;v<=cnt;v++)
 55       {
 56             if(edges[u][v]>0&&level[v]==level[u]+1&&(t=DFS(v,min(minv,edges[u][v]))))
 57             {
 58                   edges[u][v]-=t;
 59                   edges[v][u]+=t;
 60                   return t;
 61             }
 62       }
 63       level[u]=0xff;
 64       return 0;
 65 }
 66
 67 int main()
 68 {
 69       while(scanf("%d%d%d",&n,&f,&d)!=EOF)
 70       {
 71             memset(edges,0,sizeof(edges));
 72             int w;
 73             for(int i=1;i<=f;i++)
 74             {
 75                   scanf("%d",&w);
 76                   edges[0][i]=w;
 77             }
 78             // 0    1--> f,           f+1--->f+n,           f+n+1  --> f+2*n     f+2*n+1-->f+2*n+d      f+2*n+d+1;
 79             for(int i=1;i<=n;i++)
 80                   edges[f+i][f+n+i]=1;
 81             cnt=f+2*n+d+1;
 82             for(int i=1;i<=d;i++)
 83             {
 84                   scanf("%d",&w);
 85                   edges[f+2*n+i][cnt]=w;
 86             }
 87             for(int i=1;i<=n;i++)
 88             {
 89                   scanf("%s",str);
 90                   for(int j=0;j<f;j++)
 91                   {
 92                         if(str[j]=='Y')
 93                         {
 94                               edges[j+1][f+i]=1;
 95                         }
 96                   }
 97             }
 98
 99             for(int i=1;i<=n;i++)
100             {
101                   scanf("%s",str);
102                   for(int j=0;j<d;j++)
103                   {
104                         if(str[j]=='Y')
105                         {
106                               edges[f+n+i][f+2*n+j+1]=1;
107                         }
108                   }
109             }
110             int ans=0;
111             int t;
112             while(BFS())
113             {
114                   while(t=DFS(0,INF))
115                   {
116                         ans+=t;
117                   }
118             }
119             printf("%d\n",ans);
120       }
121       return 0;
122 }

time  124ms

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <string>
  7 #include <vector>
  8 #include <set>
  9 #include <map>
 10 #include <stack>
 11 #include <queue>
 12 #include <sstream>
 13 #include <iomanip>
 14 using namespace std;
 15 typedef long long LL;
 16 const int INF = 0x4fffffff;
 17 const double EXP = 1e-5;
 18 const int MS = 805;
 19 const int SIZE = 200005;
 20
 21
 22 struct edge
 23 {
 24     int v,w,next;
 25 }edges[SIZE];
 26
 27 int head[MS];
 28 int level[MS];
 29 int que[MS];
 30 int qs,qe,cnt;
 31 int n,f,d;
 32 int src,des;
 33
 34 void add(int u,int v,int w)
 35 {
 36     edges[cnt].v=v;edges[cnt].w=w;edges[cnt].next=head[u];head[u]=cnt++;
 37     edges[cnt].v=u;edges[cnt].w=0;edges[cnt].next=head[v];head[v]=cnt++;
 38 }
 39
 40 int BFS()
 41 {
 42     memset(level,-1,sizeof(level));
 43     qs=qe=0;
 44     level[src]=0;
 45     que[qe++]=src;
 46     while(qs<qe)
 47     {
 48         int u=que[qs++];
 49         for(int i=head[u];i!=-1;i=edges[i].next)
 50         {
 51             int v=edges[i].v;
 52             if(level[v]<0&&edges[i].w>0)
 53             {
 54                 level[v]=level[u]+1;
 55                 que[qe++]=v;
 56             }
 57         }
 58     }
 59     return level[des]>0;
 60 }
 61
 62 int DFS(int u,int minv)
 63 {
 64     if(u==des)
 65         return minv;
 66     int t;
 67     for(int i=head[u];i!=-1;i=edges[i].next)
 68     {
 69         int v=edges[i].v;
 70         if(edges[i].w>0&&level[v]==level[u]+1&&(t=DFS(v,min(edges[i].w,minv))))
 71         {
 72             edges[i].w-=t;
 73             edges[i^1].w+=t;
 74             return t;
 75         }
 76     }
 77     level[u]=-1;  // 从u出发无法找到argument
 78     return 0;
 79 }
 80
 81
 82 int main()
 83 {
 84       while(scanf("%d%d%d",&n,&f,&d)!=EOF)
 85       {
 86             memset(head,-1,sizeof(head));
 87             int w;
 88             char str[MS];
 89             des=f+2*n+d+1;
 90             src=cnt=0;
 91             for(int i=1;i<=f;i++)
 92             {
 93                   scanf("%d",&w);
 94                  add(src,i,w);
 95             }
 96             // 0    1--> f,           f+1--->f+n,           f+n+1  --> f+2*n     f+2*n+1-->f+2*n+d      f+2*n+d+1;
 97             for(int i=1;i<=n;i++)
 98                   add(f+i,f+n+i,1);
 99             for(int i=1;i<=d;i++)
100             {
101                   scanf("%d",&w);
102                  add(f+2*n+i,des,w);
103             }
104             for(int i=1;i<=n;i++)
105             {
106                   scanf("%s",str);
107                   for(int j=0;j<f;j++)
108                   {
109                         if(str[j]=='Y')
110                         {
111                               add(j+1,f+i,1);
112                         }
113                   }
114             }
115
116             for(int i=1;i<=n;i++)
117             {
118                   scanf("%s",str);
119                   for(int j=0;j<d;j++)
120                   {
121                         if(str[j]=='Y')
122                         {
123                               add(f+n+i,f+2*n+j+1,1);
124                         }
125                   }
126             }
127             int ans=0;
128             int t;
129             while(BFS())
130             {
131                 while(t=DFS(src,INF))
132                 {
133                     ans+=t;
134                 }
135             }
136             printf("%d\n",ans);
137       }
138       return 0;
139 }

这题和poj  3281基本上是一样的。链接：Dining

Dining是每种食物和饮料只能分配给一个人。所以可以把容量看成1. 可以看成是一样的题目。  DINIC（）  0ms;

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <string>
  7 #include <vector>
  8 #include <set>
  9 #include <map>
 10 #include <stack>
 11 #include <queue>
 12 #include <sstream>
 13 #include <iomanip>
 14 using namespace std;
 15 typedef long long LL;
 16 const int INF = 0x4fffffff;
 17 const double EXP = 1e-5;
 18 const int MS = 805;
 19 const int SIZE = 200005;
 20
 21
 22 struct edge
 23 {
 24     int v,w,next;
 25 }edges[SIZE];
 26
 27 int head[MS];
 28 int level[MS];
 29 int que[MS];
 30 int qs,qe,cnt;
 31 int n,f,d;
 32 int src,des;
 33
 34 void add(int u,int v,int w)
 35 {
 36     edges[cnt].v=v;edges[cnt].w=w;edges[cnt].next=head[u];head[u]=cnt++;
 37     edges[cnt].v=u;edges[cnt].w=0;edges[cnt].next=head[v];head[v]=cnt++;
 38 }
 39
 40 int BFS()
 41 {
 42     memset(level,-1,sizeof(level));
 43     qs=qe=0;
 44     level[src]=0;
 45     que[qe++]=src;
 46     while(qs<qe)
 47     {
 48         int u=que[qs++];
 49         for(int i=head[u];i!=-1;i=edges[i].next)
 50         {
 51             int v=edges[i].v;
 52             if(level[v]<0&&edges[i].w>0)
 53             {
 54                 level[v]=level[u]+1;
 55                 que[qe++]=v;
 56             }
 57         }
 58     }
 59     return level[des]>0;
 60 }
 61
 62 int DFS(int u,int minv)
 63 {
 64     if(u==des)
 65         return minv;
 66     int t;
 67     for(int i=head[u];i!=-1;i=edges[i].next)
 68     {
 69         int v=edges[i].v;
 70         if(edges[i].w>0&&level[v]==level[u]+1&&(t=DFS(v,min(edges[i].w,minv))))
 71         {
 72             edges[i].w-=t;
 73             edges[i^1].w+=t;
 74             return t;
 75         }
 76     }
 77     level[u]=-1;  // 从u出发无法找到argument
 78     return 0;
 79 }
 80
 81
 82 int main()
 83 {
 84       while(scanf("%d%d%d",&n,&f,&d)!=EOF)
 85       {
 86             memset(head,-1,sizeof(head));
 87             int w;
 88             char str[MS];
 89             des=f+2*n+d+1;
 90             src=cnt=0;
 91             for(int i=1;i<=f;i++)
 92                  add(src,i,1);
 93             // 0    1--> f,           f+1--->f+n,           f+n+1  --> f+2*n     f+2*n+1-->f+2*n+d      f+2*n+d+1;
 94             for(int i=1;i<=n;i++)
 95                   add(f+i,f+n+i,1);
 96             for(int i=1;i<=d;i++)
 97                  add(f+2*n+i,des,1);
 98             int fsum,dsum,t;
 99             for(int i=1;i<=n;i++)
100             {
101                 scanf("%d%d",&fsum,&dsum);
102                 for(int j=1;j<=fsum;j++)
103                 {
104                     scanf("%d",&t);
105                     add(t,f+i,1);
106                 }
107                 for(int j=1;j<=dsum;j++)
108                 {
109                     scanf("%d",&t);
110                     add(f+n+i,f+2*n+t,1);
111                 }
112             }
113             int ans=0;
114             while(BFS())
115             {
116                 while(t=DFS(src,INF))
117                 {
118                     ans+=t;
119                 }
120             }
121             printf("%d\n",ans);
122       }
123       return 0;
124 }