[缩点] poj 1236 Network of Schools

/**
[缩点] poj 1236 Network of Schools
A:找一个最小点集,从这个点集出发可以到达图上每个点
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),只有一个强连通分量时为0

*/
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;

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

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
*/