PIGS
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 21917 | Accepted: 10034 |
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
Source
大致题意：
有 M 个猪圈，每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依
次来了 N 个顾客，每个顾客分别会打开指定的几个猪圈，从中买若干头猪。每
个顾客分别都有他能够买的数量的上限。每个顾客走后，他打开的那些猪圈中的
猪，都可以被任意地调换到其它开着的猪圈里，然后所有猪圈重新关上。问总共
最多能卖出多少头猪。（1 <= N <= 100, 1 <= M <= 1000）
举个例子来说。有 3 个猪圈，初始时分别有 3、 1 和 10 头猪。依次来了 3 个顾客，
第一个打开 1 号和 2 号猪圈，最多买 2 头；第二个打开 1 号和 3 号猪圈，最多买
3 头；第三个打开 2 号猪圈，最多买 6 头。那么，最好的可能性之一就是第一个
顾客从 1 号圈买 2 头，然后把 1 号圈剩下的 1 头放到 2 号圈；第二个顾客从 3
号圈买 3 头；第三个顾客从 2 号圈买 2 头。总共卖出 2+3+2=7 头。
分析：直观地建图，合并点，得到更优的建图方式.具体可以参看Edelweiss的网络流建模汇总这篇文章.
有 M 个猪圈，每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依
次来了 N 个顾客，每个顾客分别会打开指定的几个猪圈，从中买若干头猪。每
个顾客分别都有他能够买的数量的上限。每个顾客走后，他打开的那些猪圈中的
猪，都可以被任意地调换到其它开着的猪圈里，然后所有猪圈重新关上。问总共
最多能卖出多少头猪。（1 <= N <= 100, 1 <= M <= 1000）
举个例子来说。有 3 个猪圈，初始时分别有 3、 1 和 10 头猪。依次来了 3 个顾客，
第一个打开 1 号和 2 号猪圈，最多买 2 头；第二个打开 1 号和 3 号猪圈，最多买
3 头；第三个打开 2 号猪圈，最多买 6 头。那么，最好的可能性之一就是第一个
顾客从 1 号圈买 2 头，然后把 1 号圈剩下的 1 头放到 2 号圈；第二个顾客从 3
号圈买 3 头；第三个顾客从 2 号圈买 2 头。总共卖出 2+3+2=7 头。
分析：直观地建图，合并点，得到更优的建图方式.具体可以参看Edelweiss的网络流建模汇总这篇文章.
#include <cstdio> #include <queue> #include <vector> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 120,maxm = 2020,inf = 0x7fffffff; int S,T,head[maxn],to[maxm],nextt[maxm],tot = 2,c[maxm],w[maxm]; int n,m,flag[maxm],ans,vis[maxn]; vector <int> G[maxn]; void add(int x,int y,int z) { w[tot] = z; to[tot] = y; nextt[tot] = head[x]; head[x] = tot++; } bool bfs() { queue <int> q; memset(vis,-1,sizeof(vis)); vis[S] = 0; q.push(S); while (!q.empty()) { int u = q.front(); q.pop(); if (u == T) return true; for (int i = head[u];i;i = nextt[i]) { int v = to[i]; if (w[i] && vis[v] == -1) { vis[v] = vis[u] + 1; q.push(v); } } } return false; } int dfs(int u,int f) { if (u == T) return f; int res = 0; for (int i = head[u];i;i = nextt[i]) { int v = to[i]; if (vis[v] == vis[u] + 1 && w[i]) { int temp = dfs(v,min(f - res,w[i])); res += temp; w[i] -= temp; w[i ^ 1] += temp; if (res == f) return res; } } if (!res) vis[u] = -1; return res; } void dinic() { while (bfs()) ans += dfs(S,inf); } int main() { scanf("%d%d",&m,&n); S = 0; T = n + 1; for (int i = 1; i <= m; i++) scanf("%d",&c[i]); for (int i = 1; i <= n; i++) { int num; scanf("%d",&num); for (int j = 1; j <= num; j++) { int x; scanf("%d",&x); G[i].push_back(x); } scanf("%d",&num); add(i,T,num); add(T,i,0); } for (int i = 1; i <= n; i++) { for (int j = 0;j < G[i].size(); j++) { int v = G[i][j]; if (!flag[v]) { flag[v] = i; add(S,i,c[v]); add(i,S,0); } else { add(flag[v],i,inf); add(i,flag[v],0); flag[v] = i; } } } dinic(); printf("%d ",ans); return 0; }