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POJ1149

  1 #include <iostream>
  2 #include <cstdlib>
  3 #include <cstring>
  4 #include <cstdio>
  5 #include <queue>
  6 const int inf = 1<<30 , N = 100 + 11 , M = 1000 + 11 ;
  7 using namespace std ;
  8 int m , n  ;
  9 int head[N] , tot ;
 10 struct id
 11 {
 12     int to , nxt , val ;
 13 } edge[(N*N)<<1] ;
 14 int dis[N*M] , cur[N*M] , num[M] , fir[M] , edg[N];
 15 bool vis[M] , use[N][N];  
 16 queue< int > Q ;
 17 
 18 inline void  Init( )
 19 {
 20     freopen( "NSOOJ10745.in" , "r" , stdin ) ;
 21     freopen( "NSOOJ10745.out" , "w" , stdout ) ;
 22 }
 23 
 24 
 25 int read(  )
 26 {
 27     char ch = getchar(  ) ; int k = 1 , ret = 0 ;
 28     while( ch > '9' || ch < '0' ) { if( ch == '-' ) k = -1 ; ch = getchar( ) ; }
 29     while( ch <= '9' && ch >= '0' ) ret = ret * 10 + ch - '0' , ch = getchar( ) ;
 30     return ret * k ;
 31 }
 32 
 33 
 34 void add( int u , int v , int c )
 35 {
 36     edge[++tot].to = v ; edge[tot].nxt = head[u] ;
 37     edge[tot].val = c ; head[u] = tot ; 
 38 }
 39 
 40 void input(  )
 41 {
 42     m = read( ) , n = read( ) ; tot = 1 ;
 43     memset( head , -1 , sizeof(head) ) ;
 44     for( int x = 1 ; x <= m ; ++x )
 45         num[x] = read( ) ;
 46     int a , b  ;
 47     for( int x = 1 ;  x <= n ; ++x )
 48     {
 49         a = read(  ) ;
 50         for( int y = 1 ; y <= a ; ++y )
 51         {
 52             b = read(  ) ;
 53             if( !vis[b] )
 54             {
 55                 if( !edg[x] ) 
 56                 {
 57                     add( 0 , x , num[b] ) ;
 58                     edg[x] = tot ;
 59                     add( x , 0 , 0 ) ;
 60                 }
 61                 else edge[edg[x]].val += num[b] ;
 62                 fir[b] = x ;
 63                 vis[b] = true ;
 64             }
 65             else
 66             {
 67                 if( use[fir[b]][x] ) continue ;
 68                 add( fir[b] , x , inf ) ;
 69                 add( x , fir[b] , 0 ) ;
 70                 use[fir[b]][x] = true ;
 71             }
 72         }
 73         b = read( ) ;
 74         add( x , n + 1 , b ) ;
 75         add( n+1 , x , 0 ) ;
 76     }
 77 }
 78 
 79 
 80 bool bfs( ) 
 81 {
 82     memset( dis , -1 , sizeof(dis) ) ;
 83     Q.push( 0 ) ; dis[0] = 0 ;
 84     while( !Q.empty( ) ) 
 85     {
 86         int u = Q.front( ) ; Q.pop( ) ;
 87         for( int i = head[u] ; ~i ; i = edge[i].nxt )
 88         {
 89             int v = edge[i].to ;
 90             if( dis[v] < 0 && edge[i].val > 0 )
 91             {
 92                 dis[v] = dis[u] + 1 ;
 93                 Q.push( v ) ;
 94             }
 95             
 96         }
 97     }
 98     if( dis[n+1] > 0 ) return true ;
 99     return false ;
100 }
101  
102  
103 int dfs( int u , int inf )
104 {
105     if( u == n + 1 ) return inf ;
106     int ans = 0 , cost = 0 ;
107     for( int i = cur[u] ; ~i ; i = edge[i].nxt )
108     {
109         int v = edge[i].to ;
110         if( dis[v] != dis[u] + 1 ) continue ;
111         ans = dfs( v , min( inf - cost , edge[i].val ) ) ;
112         edge[i].val -= ans ; edge[i^1].val += ans ;
113         if( edge[i].val ) cur[u] = i ;
114         cost += ans ;
115         if( cost == inf ) return inf ;
116     }
117     if( !cost ) dis[u] = -1 ;
118     return cost ;
119 }
120  
121  
122 
123 void sov(  )
124 {
125     int ans = 0;
126     while( bfs( ) )
127     {
128         for( int x = 0 ; x <= n + 1 ; ++x ) cur[x] = head[x] ;
129         ans += dfs( 0 , inf ) ;
130     }
131     printf( "%d
" , ans ) ;
132 
133 }
134 
135 
136 int main(  )
137 {
138 //    Init(  ) ;
139     input(  ) ;
140     sov( ) ;
141 //    fclose( stdin ) ;
142 //    fclose( stdout ) ;
143     return 0 ;
144     
145 }

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

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原文地址：https://www.cnblogs.com/Ateisti/p/5921968.html