Network of Schools poj 1236

持续更新中。。。。

Network of Schools

http://poj.org/problem?id=1236

给你一个有向图，有环的

1.有个超级源点s，s最少连几个点时，可以从s走到图上任意一点

2.最少添加几条边可以把图变成一个强连通分量

ans1---缩点之后,入度为0 的点的个数

ans2----缩点后,max(入度为0的点,出度为0的点),因为可以挨个连接,连的边转一圈,建议自己画一画

 

记得特判一下

 

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn = 2e5+11;
vector<int> G[maxn];
void add(int x,int y){
   G[x].push_back(y);
}

int n;
int ans,clor[maxn],df;
int dfn[maxn],low[maxn];
stack<int>s;


int tarjan(int x){
    s.push(x);
    dfn[x] = low[x] = ++df;
    for (int i = 0; i < G[x].size();i++){
        int p = G[x][i];
        if(!dfn[p]){
            tarjan(p);
            low[x] = min(low[x], low[p]);
        }
        else if(!clor[p]){
            low[x] = min(low[x], dfn[p]);
        }
    }

    if(low[x] == dfn[x]){
        ans++;
        while(1){
            int a = s.top();
            s.pop();
            clor[a] = ans;
            if(a == x) break;
        }
    }
    return 0;
}

int in[maxn], ot[maxn];

int main(){
    cin >> n;
    int x;
    for (int i = 1; i <= n;i++){
      
        while(1){
            cin >> x;
            if(x == 0) break;
            add(i, x);
        }
    }
    for (int i = 1; i <= n;i++){
        if(dfn[i] == 0){
            tarjan(i);
        }
    }
    for (int i = 1; i <= n;i++){
        for (int j = 0; j < G[i].size();j++){
            int p = G[i][j];
            int x = clor[i];
            int y = clor[p];
            if(x != y){
                //x---->y
                in[y]++;
                ot[x]++;
            }
        }
    }

    int a = 0, b = 0;
    for (int i = 1; i <= ans; i++){
        if(in[i] == 0) a++;
        if(ot[i] == 0) b++;
    }
    if(ans == 1){
        cout << 1 << endl;
        cout << 0 << endl;
        return 0;
    }

    cout << a << endl;
    cout << max(a,b) << endl;
    return 0;
}