tarjan 算法模板

做了一些树形DP实在不想做了看到von在玩tarjan算法，好像放假之前看过关于rmq 以及tarjan算法的说明只是没具体弄明白。今天看了一下von分享的资料不错。找了一个模板题就当tarjan算法的模板吧

http://acm.hdu.edu.cn/showproblem.php?pid=1269

判断是否为强连通图 bcnt == 1只有一个强连通分量且为自己 bcnt >1 存在多个强连通分量

View Code

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#define maxn 10007
using namespace std;
vector<int>g[maxn];
int DFN[maxn],LOW[maxn],Belong[maxn],stack[maxn];
bool instack[maxn];
int n,m;
int bcnt,top,index;
void tarjan(int i)
{
    int j;
    DFN[i] = LOW[i] = ++index;
    stack[++top] = i;
    instack[i] = true;
    int size = g[i].size();
    for (int k = 0; k < size; ++k)
    {
        j = g[i][k];
        if(!DFN[j])
        {
            tarjan(j);
            if(LOW[j] < LOW[i])
            LOW[i] = LOW[j];
        }
        else if(instack[j] && DFN[j] < LOW[i])
        {
            LOW[i] = DFN[j];
        }
    }
    if(DFN[i] == LOW[i])
    {
        bcnt ++;
        do
        {
             j = stack[top--];
             instack[j] = false;
             Belong[j] = bcnt;
        }while(j != i);
    }
}
void solve()
{
    int i;
    bcnt = top = index = 0;
    for (i = 1; i <= n; ++i)
    {
        if(!DFN[i])
        tarjan(i);
    }
    if(bcnt == 1)
    printf("Yes\n");
    else
    printf("No\n");
}
int main()
{
    int i,x,y;
    while(cin>>n>>m)
    {
        if(!n && !m) break;
        for (i = 0; i <= n; ++i)
        {
            DFN[i] = LOW[i] = Belong[i] = 0;
            instack[i] = false;
            g[i].clear();
        }
        for (i = 0; i < m; ++i)
        {
            cin>>x>>y;
            g[x].push_back(y);
        }
        solve();
        /*for (i = 1; i <= n; ++i)
        {
            printf("%d ",Belong[i]);
        }
        printf("\n");*/
    }
    return 0;
}