[缩点] poj 1236 Network of Schools

/**
[缩点] poj 1236 Network of Schools
A:找一个最小点集,从这个点集出发能够到达图上每一个点
B:至少添加多少条边使这个有向图变成强连通图
首先强连通缩点统计出度为0的结点数为x，入度为0的结点数为y

问题A为 y
问题B为 max(x,y),只有一个强连通份量时为0 
 
*/ 
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;

#define N 124
int low[N],pre[N],id[N],cnt,scnt,sp,stk[N],out[N],in[N],n;
vector<int> g[N];

void tarjan(int u){
	int t,minc = low[u] = pre[u] = cnt ++;
	stk[sp++] = u;
	for(int i = 0; i < g[u].size(); ++i){
		int v = g[u][i];
		if(-1 == pre[v])
			tarjan(v);
		if(minc > low[v])
			minc = low[v];
	}
	if(minc < low[u]){
		low[u] = minc;
		return ;
	}
	do{
		id[t = stk[--sp] ] = scnt;
		low[t] = n + 1;
	}while(t != u);
	++scnt;
}
int main(){
	int i,j,k;
	//while(scanf("%d",&n)!= EOF){
		scanf("%d",&n);{
		memset(pre,-1,sizeof(pre));
		memset(g,0,sizeof(g));
		sp = cnt = scnt = 0;
		for(i = 1; i <= n; ++i){
			while(scanf("%d",&j) && j)
				g[i].push_back(j);
		}
		for(i = 1; i <= n; ++i)
			if(-1 == pre[i])
				tarjan(i);
		for(i = 1; i <= n; ++i)
			for(j = 0; j < g[i].size(); ++j){
				k = g[i][j];
				if(id[i] != id[k])
					++in[id[k]],++out[id[i]];
			}
		for(j = k = i = 0; i < scnt; ++i){
			j += !in[i];
			k += !out[i];
		}
		k = k > j ? k : j;
		k = scnt == 1 ? 0 : k;
		printf("%d\n%d\n",j,k);
	}
	return 0 ;
}
/**
5
3 4 0
4 5 0
0
0 
0
*/