POJ1149 PIGS[dinic()最大流]

PIGS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12578		Accepted: 5560

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.
An unlimited number of pigs can be placed in every pig-house.
Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):
A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

Sample Output

Source

Croatia OI 2002 Final Exam - First day

这题主要是构图比较难，这题以人为节点，如果有公共节点的人就作为前驱

code:

  1 #include<iostream>
  2 using namespace std;
  3 
  4 #define MAXN 110
  5 #define inf 0x7fffffff
  6 
  7 int n,m;
  8 int cnt[1010];
  9 int pre[1010];
 10 int dis[1010];
 11 int edgenum;
 12 
 13 typedef struct edge
 14 {
 15     int from,to,cap;
 16     int next;
 17 }Edge;
 18 Edge edge[MAXN*10];
 19 int head[MAXN];
 20 
 21 void addedge(int a,int b,int c)
 22 {
 23     edge[edgenum].from=a,edge[edgenum].to=b,edge[edgenum].cap=c,edge[edgenum].next=head[a],head[a]=edgenum++;
 24     edge[edgenum].from=b,edge[edgenum].to=a,edge[edgenum].cap=0,edge[edgenum].next=head[b],head[b]=edgenum++;
 25 }
 26 
 27 int dinic(int st,int et)
 28 {
 29     int sta[1010];
 30     int que[1010];
 31     int ans=0;
 32     while(true)
 33     {
 34         memset(dis,-1,sizeof(dis));
 35         int front=0,tail=0;
 36         int u,v;
 37         dis[st]=0;
 38         que[tail++]=st;
 39         while(front<tail)
 40         {
 41             v=que[front++];
 42             for(int i=head[v];i!=-1;i=edge[i].next)
 43             {
 44                 u=edge[i].to;
 45                 if(dis[u]==-1&&edge[i].cap>0)
 46                 {
 47                     dis[u]=dis[v]+1;
 48                     que[tail++]=u;
 49                     if(u==et)
 50                     {
 51                         front=tail;
 52                         break;
 53                     }
 54                 }
 55             }
 56         }
 57         if(dis[et]==-1)
 58             break;
 59         int t=0;
 60         int s=st;
 61         while(true)
 62         {
 63             if(s!=et)
 64             {
 65                 int i;
 66                 for(i=head[s];i!=-1;i=edge[i].next)
 67                     if(edge[i].cap>0&&dis[edge[i].from]+1==dis[edge[i].to])
 68                         break;
 69                 if(i!=-1)
 70                 {
 71                     sta[t++]=i;
 72                     s=edge[i].to;
 73                 }
 74                 else
 75                 {
 76                     if(t==0)
 77                         break;
 78                     dis[edge[sta[--t]].to]=-1;
 79                     s=edge[sta[t]].from;
 80                 }
 81             }
 82             else
 83             {
 84                 int mindis=inf;
 85                 int tag;
 86                 for(int i=0;i<t;i++)
 87                     if(mindis>edge[sta[i]].cap)
 88                     {
 89                         mindis=edge[sta[i]].cap;
 90                         tag=i;
 91                     }
 92                 ans+=mindis;
 93                 for(int i=0;i<t;i++)
 94                 {
 95                     edge[sta[i]].cap-=mindis;
 96                     edge[sta[i]^1].cap+=mindis;
 97                 }
 98                 s=edge[sta[tag]].from;
 99                 t=tag;
100             }
101         }
102     }
103     return ans;
104 }
105 
106 int main()
107 {
108     int i,j;
109     int vst[1010];
110     edgenum=0;
111     memset(vst,0,sizeof(vst));
112     memset(head,-1,sizeof(head));
113     memset(pre,0,sizeof(pre));
114     scanf("%d%d",&m,&n);                   //m：猪栏数量；n：人数
115     int s=0,t=n+1;
116     for(int i=1;i<=m;i++)
117         scanf("%d",&cnt[i]);
118     for(i=1;i<=n;i++)
119     {
120         int count,a,cap;
121         int ans=0;
122         scanf("%d",&count);
123         for(j=0;j<count;j++)
124         {
125             scanf("%d",&a);
126             if(!vst[a])
127             {
128                 ans+=cnt[a];
129                 vst[a]=true;
130                 pre[a]=i;
131             }
132             else
133             {
134                 addedge(pre[a],i,inf);
135             }            
136         }
137         if(ans)
138             addedge(s,i,ans);
139         scanf("%d",&cap);
140         addedge(i,t,cap);
141     }
142     printf("%d\n",dinic(s,t));
143     return 0;
144 }