[强连通分支]poj_1236_network of schools

题意：给定一个有向图，求最少要从几个结点开始能把图遍历完，最少添加多少条边，使得整个图强连通。

先缩点，然后对于缩点以后的图，入度为0的点的个数就是第一问的解，第二问的解是入度为0结点个数和出度为0结点个数中的最大值，第二问参考的解题报告是：http://hi.baidu.com/oichampion/blog/item/1882abd7d86adec5a9ec9aa5.html

代码:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<stack>
using namespace std;
#define MIN(a,b) a>b?b:a
#define MAX(a,b) a<b?b:a
const int MAXN = 110;
int sg[MAXN][MAXN],vis[MAXN],dfn[MAXN],low[MAXN],belong[MAXN],indeg[MAXN],outdeg[MAXN],scnt,deep;
vector<int> g[MAXN];
stack<int> S;
void init()
{
memset(belong,-1,sizeof(belong));
memset(vis,0,sizeof(vis));
memset(sg,0,sizeof(sg));
memset(indeg,0,sizeof(indeg));
memset(outdeg,0,sizeof(outdeg));
for(int i=0;i<MAXN;i++)g[i].clear();
while(!S.empty())S.pop();scnt=0;deep=0;
}
void tarjan(int u,int &deep)
{
low[u] = dfn[u] = deep;
vis[u] = 1;S.push(u);deep++;
for(int i=0;i<g[u].size();i++)
{
int v = g[u][i];
if(!vis[v])
{
tarjan(v,deep);
low[u] = MIN(low[u],low[v]);
}
else if(belong[v]==-1)
{
low[u] = MIN(low[u],dfn[v]);
}
}
if(low[u]==dfn[u])
{
int v;
do
{
v = S.top();
S.pop();
belong[v] = scnt;
}while(v!=u);
scnt++;
}
}
void getscc(int n)
{
for(int i=1;i<=n;i++)
{
if(!vis[i])tarjan(i,deep);
}
}
void getsg(int n)//获得缩点图
{
for(int i=1;i<=n;i++)
{
for(int j=0;j<g[i].size();j++)
{
if(belong[i]!=belong[g[i][j]])
{
sg[belong[i]][belong[g[i][j]]]=1;
}
}
}
}
void cacsgindeg()//计算缩点图的入度
{
for(int i=0;i<scnt;i++)
{
for(int j=0;j<scnt;j++)
{
if(sg[j][i])indeg[i]++;
if(sg[i][j])outdeg[i]++;
}
}
}
int getzerodeg()
{
int cnt = 0;
for(int i=0;i<scnt;i++)
{
if(indeg[i]==0)cnt++;
}
return cnt;
}
int getzeroout()
{
int cnt = 0;
for(int i=0;i<scnt;i++)
{
if(outdeg[i]==0)cnt++;
}
return cnt;
}

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int u;init();
for(int i=1;i<=n;i++)
{
while(scanf("%d",&u),u)
{
g[i].push_back(u);
}
}
getscc(n);
getsg(n);
cacsgindeg();
int copy = getzerodeg(),edge = MAX(copy,getzeroout());
if(scnt!=1){
printf("%d\n%d\n",copy,edge);
}else printf("1\n0\n");
}
return 0;
}