标签:style io ar color sp for on bs ef
强连通分量压缩是
先缩点,然后计算各个强连通分量的入度为0的个数,出度为0的个数求他们最大值
#include <stdio.h>
#include <string.h>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
#define N 20005
stack<int>sta;
vector<int>mp[N];
int dfn[N];
int low[N];
int InStack[N];
int indexx,number;
int n, m;
int id[N];
void tarjan(int u)
{
InStack[u] = 1;
low[u] = dfn[u] = ++ indexx;
sta.push(u);
for (int i = 0; i < mp[u].size(); ++ i)
{
int t = mp[u][i];
if (dfn[t] == 0)
{
tarjan(t);
low[u] = min(low[u], low[t]);
}
else if (InStack[t] == 1)
{
low[u] = min(low[u], dfn[t]);
}
}
if (low[u] == dfn[u])
{
++ number;
while (!sta.empty())
{
int v = sta.top();
sta.pop();
id[v]=number;
InStack[v] = 0;
if (v == u)
break;
}
}
}
int main()
{
//freopen("input.txt","r",stdin);
int T;
while(scanf("%d",&T)==1)
{
while(T--)
{
scanf("%d%d",&n,&m);
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(InStack, 0, sizeof(InStack));
indexx = number = 0;
for (int i = 1; i <= n; ++ i)
{
mp[i].clear();
}
while(!sta.empty())
sta.pop();
for(int i=1; i<=m; i++)
{
int a,b;
scanf("%d%d",&a,&b);
mp[a].push_back(b);
}
for(int i=1; i<=n; i++)
if(!dfn[i])
tarjan(i);
if(number==1)
{
printf("0\n");
continue;
}
memset(dfn, 0, sizeof( dfn));
memset(low, 0, sizeof (low));
for(int i=1; i<=n; i++)//计算出入度
{
for(int j=0; j<mp[i].size(); j++)
{
if(id[i]!=id[mp[i][j]]) dfn[id[i]]++, low[id[mp[i][j]]]++;//如果他们不在同一个的强连通分量话,把id【i】的出度++,id【他的节点】所在的连通分量的入度加加
}
}
/*for(int i=1; i<=number; i++){
printf("%d %d\n",dfn[i],low[i]);
}*/
int ans1=0, ans2=0;
for(int i=1; i<=number; i++)
{
if(dfn[i]==0) ans1++;
if(low[i]==0) ans2++;
}
printf("%d\n", max(ans1, ans2));
}
}
}hdu 2767Proving Equivalences(强连通分量压缩 )
标签:style io ar color sp for on bs ef
原文地址:http://blog.csdn.net/u013076044/article/details/41697555