縮點 agreement iostream send push nbsp ima lis stream

Network of Schools

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24880		Accepted: 9900

題目鏈接：http://poj.org/problem?id=1236

Description:

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.

Input:

The first line contains an integer N: the number of schools in the network (2 <= N <= 100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.

Output:

Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.

Sample Input:

5
2 4 3 0
4 5 0
0
0
1 0

Sample Output:

1
2

題意：

給出一個有向圖，然後要你輸出兩個任務的答案：

1.至少需要從多少個點出發，能夠到達所有的點；2.最少需要連多少條邊，能夠使得從任意點出發都能夠到達其它所有點。

題解：

這個題我一開始想的就是直接暴力，但很明顯第二個問題行不通，所以就要考慮一些性質，或者用一些數學思想。

第一個問題還是比較好想，入度為0的點的個數即位答案，如果不存在入度為0的點，答案就是1。簡略證明如下(題目保證圖是連通的)：

假設入度為0的點為n，那麽至少需要n個點才能遍及所有點，然後對於其余入度非0的點來說，必然是由其他點到達的，如果這個點不在環上，那麽就必定是從一個入度為0的點來的；如果這個點在環上，這個環中的所有點也會由其余入度為0的點到達；假設這是個單獨的環，那麽答案為1。

第二個問題要求所有點都互相可以到達。那麽我們可以知道的是，圖中必然不會存在入度為0以及出度為0的點，假設這兩者的個數分別為n,m。

那麽最優的連邊方法就是入度為0的點與出度為0的點匹配，最後剩下的亂連就行了，所以最後答案就是max(n,m)。證明的話yy一下吧。

因為我們剛才是基於有向無環圖來思考的，環的存在應該把它當作一個點，所以考慮Tarjan縮波點就行了。

代碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
typedef long long ll;
const int N = 105;
int n,tot;
int head[N],in[N],out[N],low[N],dfn[N],vis[N],scc[N];
struct Edge{
    int u,v,next;
}e[N*N<<1],edge[N*N<<1];
void adde(int u,int v){
    e[tot].u=u;e[tot].v=v;e[tot].next=head[u];head[u]=tot++;
}
stack <int> s;
int T,num;
void Tarjan(int u){
    dfn[u]=low[u]=++T;vis[u]=1;
    s.push(u);
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(!vis[v]){
            Tarjan(v);
            low[u]=min(low[u],low[v]);
        }else if(!scc[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]){
        num++;int now;
        do{
            now = s.top();s.pop();
            scc[now]=num;
        }while(!s.empty() && now!=u);
    }
}
int main(){
    scanf("%d",&n);
    int m=0;
    memset(head,-1,sizeof(head));
    for(int i=1;i<=n;i++){
        int v;
        while(scanf("%d",&v)!=EOF){
            if(v==0) break ;
            edge[++m].u=i;edge[m].v=v;
            adde(i,v);
        }
    }
    //cout<<m<<endl;
    for(int i=1;i<=n;i++){
        if(!vis[i]) Tarjan(i);
    }
    for(int i=1;i<=m;i++){
        int u=edge[i].u,v=edge[i].v;
        if(scc[u]!=scc[v]){
            in[scc[v]]++;out[scc[u]]++;
        }
    }
    int cnt1=0,cnt2=0;
    for(int i=1;i<=num;i++){
        if(!in[i]) cnt1++;
        if(!out[i]) cnt2++;
    }
    //cout<<num<<endl;
    if(num==1) cout<<1<<endl<<0;
    else cout<<cnt1<<endl<<max(cnt2,cnt1);
    return 0;
}

POJ1236：Network of Schools (思維+Tarjan縮點)