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
3 3 3 1 10 2 1 2 2 2 1 3 3 1 2 6Sample Output
7分析主要是建图，增加一个源点和汇点，将源点和第一个有这个猪圈钥匙的顾客连起来，边权就是这个猪圈猪的数量，如果源点和某个节点有重边的话，合并就是了之后的顾客对于已经开了的猪圈，将上一个顾客和他相连，边权是正无穷，可以用一个last数组来记录上一个买这个猪圈的顾客每个顾客和汇点相连，边权是他想买猪的数量。 #include<stdio.h> #include<string.h> #include<iostream> #include<queue> using namespace std; //**************************************************** //最大流模板Edmonds_Karp算法 //初始化：G[][],st,ed //****************************************************** const int MAXN = 200+10; const int INF = 0x3fffffff; int G[MAXN][MAXN];//存边的容量，没有边的初始化为0 int path[MAXN],flow[MAXN],st,ed; int n;//点的个数，编号0~n，n包括了源点和汇点 queue<int>q; int bfs() { int i,t; while(!q.empty()) q.pop();//清空队列 memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1 path[st]=0; flow[st]=INF;//源点可以有无穷的流流进 q.push(st); while(!q.empty()){ t=q.front(); q.pop(); if(t==ed) break; for(i=0;i<=n;i++){ if(i!=st&&path[i]==-1&&G[t][i]){ flow[i]=flow[t]<G[t][i]?flow[t]:G[t][i]; q.push(i); path[i]=t; } } } if(path[ed]==-1) return -1;//即找不到汇点上去了。找不到增广路径了 return flow[ed]; } int Edmonds_Karp() { int max_flow=0; int step,now,pre; while((step=bfs())!=-1){ max_flow+=step; now=ed; while(now!=st){ pre=path[now]; G[pre][now]-=step; G[now][pre]+=step; now=pre; } } return max_flow; } int val[1100]; int last[1100]; int main() { int N,M; scanf("%d%d",&M,&N); st=0,ed=N+1,n=N+1; memset(val,0,sizeof(val)); memset(last,0,sizeof(last)); for(int i=1;i<=M;i++) scanf("%d",&val[i]); for(int i=1;i<=N;i++){ int A,B,v; scanf("%d",&A); for(int j=1;j<=A;j++){ scanf("%d",&v); if(last[v]==0){ G[st][i]+=val[v]; last[v]=i; } else{ G[last[v]][i]=INF; last[v]=i; } } scanf("%d",&B); G[i][ed]=B; } printf("%d\n",Edmonds_Karp()); return 0; }
转载于:https://www.cnblogs.com/wangdongkai/p/5616618.html
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