PIGS
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 14239 | Accepted: 6330 |
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.
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.
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
3 3 3 1 10 2 1 2 2 2 1 3 3 1 2 6
Sample Output
7
#include <cstdio> #include <cstring> #define INF 300000000 //无穷大 #define MAXM 1000 //猪圈数:1≤M≤1000 #define MAXN 100 //顾客数:1≤N≤100 int s, t; //源点,汇点 int customer[MAXN+2][MAXN+2]; //N+2个结点(包括源点,汇点)之间的容量Cij int flow[MAXN+2][MAXN+2]; //节点之间的流量Fij int i, j; //循环变量 void init( ) //初始化函数,构造网络流 { int M, N; //M是猪圈的数目,N是顾客的数目 int num; //每个顾客拥有钥匙的数目 int k; //第K个猪圈的钥匙 int house[MAXM]; //存储每个猪圈中猪的数目 int last[MAXM]; //存储每个猪圈的前一个顾客的序号 memset( last, 0, sizeof(last) ); memset( customer, 0, sizeof(customer) ); scanf( "%d%d", &M, &N ); s = 0; t = N + 1; //源点、汇点 for( i=1; i<=M; i++ ) //读入每个猪圈中猪的数目 scanf( "%d", &house[i] ); for( i=1; i<=N; i++ ) //构造网络流 { scanf( "%d", &num ); //读入每个顾客拥有钥匙的数目 for( j=0; j<num; j++ ) { scanf( "%d", &k ); //读入钥匙的序号 if( last[k]==0 ) //第i个顾客是第k个猪圈的第1个顾客 customer[s][i] = customer[s][i] + house[k]; else //last[k]!=0,表示顾客i紧跟在顾客last[k]后面打开第k个猪圈 customer[ last[k] ][i] = INF; last[k] = i; } scanf( "%d", &customer[i][t] ); //每个顾客到汇点的边,权值为顾客购买猪的数量 } } void ford( ) { //可改进路径上该顶点的前一个顶点的序号,相当于标号的第一个分量, //初始为-2表示未标号,源点的标号为-1 int prev[ MAXN+2 ]; int minflow[ MAXN+2 ]; //每个顶点的可改进量α,相当于标号的第二个分量 //采用广度优先搜索的思想遍历网络,从而对所有顶点进行标号 int queue[MAXN+2]; //相当于BFS算法中的队列 int qs, qe; //队列头位置、队列尾位置 int v; //当前检查的顶点 int p; //用于保存Cij-Fij for( i=0; i<MAXN+2; i++ ) //构造零流:从零流开始标号调整 { for( j=0; j<MAXN+2; j++ ) flow[i][j] = 0; } minflow[0] = INF; //源点标号的第二个分量为无穷大 while( 1 ) //标号法 { for( i=0; i<MAXN+2; i++ ) //每次标号前,每个顶点重新回到未标号状态 prev[i] = -2; prev[0] = -1; //源点 qs = 0; queue[qs] = 0; qe = 1; //源点(顶点0)入队列 //标号过程:如果qe>=qs(相当于队列空),标号也无法再进行下去 while( qs<qe && prev[t]==-2 ) { v = queue[qs]; qs++; //取出队列头顶点 for( i=0; i<t+1; i++ ) { //如果顶点i是顶点v的"邻接"顶点,则考虑是否对顶点i进行标号 //customer[v][i]-flow[v][i]!=0能保证顶点i是v的邻接顶点, //且能进行标号 //顶点i未标号,并且Cij-Fij>0 if( prev[i]==-2 && ( p=customer[v][i]-flow[v][i]) ) { prev[i] = v; queue[qe] = i; qe++; minflow[i] = (minflow[v]<p) ? minflow[v] : p; } } } if( prev[t] == -2 ) break; //汇点t没有标号,标号法结束 for( i=prev[t], j=t; i!=-1; j=i, i=prev[i] ) //调整过程 { flow[i][j] = flow[i][j] + minflow[t]; flow[j][i] = -flow[i][j]; } } for( i=0, p=0; i<t; i++ ) //统计进入汇点的流量,即为最大流的流量 p = p + flow[i][t]; printf( "%d ", p ); } int main( ) { init( ); ford( ); return 0; }