poj 1149 PIGS(最大流)
2014-12-13 02:14
363 查看
Language: Default PIGS
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 6 Sample Output 7 |
有顾客过来买猪 顾客拥有猪圈的钥匙 可以打开猪圈 然后买该猪圈里面的猪
猪可以在从一个猪圈移到另外一个猪圈
卖猪的想要卖出最多的猪 求做多猪的数量
建图:
设立一个源点和一个汇点
将源点与每个猪圈的第一个顾客之间建边 权值为该猪圈中猪的数量 如果出现重边 权值相加
将每个顾客与汇点之间建边 权值为顾客买猪的数量
对于同一猪圈 上一个顾客与下一个顾客之间建边 权值为INF
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <string.h> #include <string> #include <vector> #include <queue> #define eps 1e-8 #define op operator #define MOD 10009 #define MAXN 1100 #define INF 30000000 #define MEM(a,x) memset(a,x,sizeof a) #define ll __int64 using namespace std; struct Dinic { struct Edge { int from,to,cap,flow; Edge(){} Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){} }; vector<Edge> edges; vector<int> G[MAXN]; bool vis[MAXN]; int d[MAXN]; int cur[MAXN]; int n,m,s,t,maxflow; void init(int n) { this->n=n; for(int i=0;i<=n;i++) G[i].clear(); edges.clear(); } void addedge(int from,int to,int cap) { edges.push_back(Edge(from,to,cap,0)); edges.push_back(Edge(to,from,0,0)); m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool bfs() { MEM(vis,0); MEM(d,-1); queue<int> q; q.push(s); d[s]=maxflow=0; vis[s]=1; while(!q.empty()) { int u=q.front(); q.pop(); int sz=G[u].size(); for(int i=0;i<sz;i++) { Edge e=edges[G[u][i]]; if(!vis[e.to]&&e.cap>e.flow) { d[e.to]=d[u]+1; vis[e.to]=1; q.push(e.to); } } } return vis[t]; } int dfs(int u,int a) { if(u==t||a==0) return a; int sz=G[u].size(); int flow=0,f; for(int &i=cur[u];i<sz;i++) { Edge &e=edges[G[u][i]]; if(d[u]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0) { e.flow+=f; edges[G[u][i]^1].flow-=f; flow+=f; a-=f; if(a==0) break; } } return flow; } int Maxflow(int s,int t) { this->s=s; this->t=t; int flow=0; while(bfs()) { MEM(cur,0); flow+=dfs(s,INF); } return flow; } }Dic; int main() { //freopen("ceshi.txt","r",stdin); int n,m; while(scanf("%d%d",&m,&n)!=EOF) { int c[MAXN];//各个猪圈的容量 int last[MAXN];//各个猪圈上一次打开的人 MEM(last,0); for(int i=1;i<=m;i++) scanf("%d",&c[i]); Dic.init(n+1); for(int i=1;i<=n;i++) { int num; scanf("%d",&num); for(int j=0;j<num;j++) { int x; scanf("%d",&x); if(last[x]==0) { last[x]=i; Dic.addedge(0,i,c[x]); } else { // last[x]=i; Dic.addedge(last[x],i,INF); last[x]=i; } } int y; scanf("%d",&y); Dic.addedge(i,n+1,y); } printf("%d\n",Dic.Maxflow(0,n+1)); } return 0; }
//以上的代码很好的解决重边问题 出现重边 直接建边
相关文章推荐
- poj 1149 pigs(最大流)
- POJ 1149 PIGS ,最大流
- POJ-1149-PIGS(最大流 标号法)
- PIGS (poj 1149 最大流)
- poj1149 PIGS(最大流+建模)
- poj1149 PIGS --- 最大流EK
- poj1149 - PIGS(最大流)
- poj 1149 PIGS 最大流
- [POJ 1149]PIGS[最大流][建图]
- POJ1149 PIGS [最大流 建图]
- poj 1149 PIGS (网络流最大流Dinic)
- POJ 1149 PIGS (最大流)
- poj-1149-PIGS(最大流)
- POJ 1149 PIGS(最大流)
- poj 1149 PIGS 最大流 太神奇的建图方式!!!
- POJ 1149 PIGS 学会构图,最大流
- POJ 1149 PIGS (最大流)
- 【最大流】POJ-1149 PIGS
- POJ 1149 PIGS【最大流】
- poj 1149 PIGS【最大流】