zoukankan      html  css  js  c++  java
  • poj1236 Network of Schools

    A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B
    You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.

    分析:先缩点,第一个问题是in[]为0的点的个数,第二个问题是max(Σin[],Σout[]);

    证明:要形成强连通分量,那么只要解决那些出度为0的或者入度为0的点,最好的方式是出度为0的点向入度为0的点连边,但他们可能不会一一对应,所以取个max。

    code:

    #include<iostream>
    #include<cstdio>
    #include<cstring> 
    #include<algorithm>
    #include<vector>
    #include<stack>
    using namespace std;
    
    const int maxn=1e5+10;
    const int INF=2e9;
    vector<int>E[maxn];
    
    struct point{
        int x,y,r,co;
    }boom[maxn];
    int dfn[maxn],low[maxn],id,vis[maxn],ans,deg[maxn];
    int num[maxn],cnt,cost[maxn],in[maxn],out[maxn];
    stack<int>s;
    
    
    bool ju(point a,point b){
        return (1ll*(a.x-b.x)*(a.x-b.x)+1ll*(a.y-b.y)*(a.y-b.y)<=1ll*a.r*a.r);  
    }
    
    void init(){
        id=cnt=0;
        memset(vis,0,sizeof(vis));
        memset(dfn,0,sizeof(dfn));
        memset(deg,0,sizeof(deg));
        memset(num,0,sizeof(num)); 
    }
    
    void tarjan(int u){
        low[u]=dfn[u]=++id;
        s.push(u);
        vis[u]=1;
        for(int i=0;i<E[u].size();i++){
            int v=E[u][i];
            if(!dfn[v]){
                tarjan(v);
                low[u]=min(low[u],low[v]);
            }
            else if(vis[v])low[u]=min(low[u],dfn[v]);
        }
        if(low[u]==dfn[u]){
            int minco=INF;
            cnt++;
            while(1){
                int Top=s.top();
                s.pop();
                vis[Top]=0;
                num[Top]=cnt; 
                if(Top==u)break;
            }
        }
    }
    
    int main(){ 
            init();
            int n;
            cin>>n;
    		getchar();
            for(int i=1;i<=n;i++){ 
    			while(1){
    				char ch=getchar();
    				while(!(ch>='0'&&ch<='9'))ch=getchar();
    				int numm=0;
    				while (ch>='0'&&ch<='9'){
    					numm=numm*10+ch-'0';
    					ch=getchar();
    				}
    				if(!numm)break;
    				E[i].push_back(numm);	
    				if(ch=='
    ')break;		
    			}
            }  
        for(int i=1;i<=n;i++)if(!dfn[i])tarjan(i); 
        for(int i=1;i<=n;i++){
            for(int j=0;j<E[i].size();j++){
               int v=E[i][j];
                if(num[i]!=num[v]){ 
    				++out[num[i]];
    				++in[num[v]];
    			} 	
            }
        } 
    	int aa=0,bb=0;
    	for(int i=1;i<=cnt;i++){ 
    		aa+=in[i]==0;
    		bb+=out[i]==0;
    	}
    	if(cnt==1)cout<<1<<endl<<0;
    	else cout<<aa<<endl<<max(aa,bb);
        return 0;
    }
    
    
  • 相关阅读:
    [转载]android开发手册
    [转载]windows phone7 学习笔记10——生命周期/墓碑化
    [转载]Windows Phone 系列 本地数据存储
    【转载】windows phone7 学习笔记12——推送通知服务
    【转载】windows phone7 学习笔记15——Bing Maps
    [转载]WP7交互特性浅析及APP设计探究
    【】windows phone7 学习笔记11——启动器与选择器
    [转载]支持的 Windows Phone 媒体编解码器
    【转载】wp7屏幕截图代码
    【转载】windows phone7 学习笔记14——地理位置服务与反应性扩展框架
  • 原文地址:https://www.cnblogs.com/GUOGaby/p/15041587.html
Copyright © 2011-2022 走看看