HDU 1629 迷宫城堡

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

解题思路：

如果整个图只有一个连通分量输出“Yes”，其它就输出“NO”

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;

const int MAXN=100010;
const int MAXM=2000010;

struct Edge{
    int to,next;
}edges[MAXM];
int head[MAXN],tot;

void init(){
    memset(head,-1,sizeof(head));
    tot=0;
}

void addEdge(int u,int v){
    edges[tot].to=v;
    edges[tot].next=head[u];
    head[u]=tot++;
}

int DFN[MAXN],LOW[MAXN],Belong[MAXN],Stack[MAXN];
int Index,top;
int scc;
bool Instack[MAXN];
int num[MAXN];

void Tarjan(int u){
    int v;
    LOW[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;

    for(int i=head[u];i!=-1;i=edges[i].next){
        v=edges[i].to;

        if(!DFN[v]){
            Tarjan(v);
            if(LOW[u]>LOW[v]) LOW[u]=LOW[v];
        }else if(Instack[v]&&LOW[u]>DFN[v])
            LOW[u]=DFN[v];
    }

    if(LOW[u]==DFN[u]){
        scc++;

        do{
            v=Stack[--top];
            Instack[v]=false;
            num[scc]++;
            Belong[v]=scc;
        }while(v!=u);
    }
}

void solve(int N){
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));

    Index=top=scc=0;

    for(int i=1;i<=N;i++){
        if(!DFN[i])
            Tarjan(i);
    }

    if(scc>1)
        cout<<"No"<<endl;
    else
        cout<<"Yes"<<endl;
}

int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF&&(n!=0||m!=0)){
            init();
        for(int i=0;i<m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            addEdge(u,v);
        }
        solve(n);
    }

}