SDUT 2604 Thrall’s Dream （单连通的判断）

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大意：给定一些有向图间的关系问任意两点是不是可达的。

思路：Tarjan后直接看入度和出度为0的数量关系，如果大于1则肯定是不可能，相连通的。

#include<map>
#include<queue>
#include<cmath>
#include<cstdio>
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
#define LL int
#define inf 0x3f3f3f3f
#define eps 1e-8
#include<vector>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1

using namespace std;

const int Ma = 10100;
struct node{
    int to,w,next;
}q[Ma];
int head[Ma],dfn[Ma],num[Ma],du[Ma],stk[Ma],vis[Ma],low[Ma];
int cnt,top,tim,scc,out[Ma],in[Ma];

bool mp[1100][1100];
void Add(int a,int b){
    q[cnt].to = b;
    q[cnt].next = head[a];
    head[a] = cnt++;
}
void init(){
    scc =  cnt = top = 0;
    tim =  1;
    memset(head,-1,sizeof(head));
    memset(dfn,0,sizeof(dfn));
    memset(num,0,sizeof(num));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(vis,0,sizeof(vis));
    memset(low,0,sizeof(low));
    memset(mp,false,sizeof(mp));
}
void Tarjan(int u){
    low[u] = dfn[u] = tim++;
    vis[u] = 1;
    stk[top++] = u;
    for(int i = head[u]; ~i ; i = q[i].next){
        int v = q[i].to;
        if(!vis[v]){
            Tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else
            low[u] = min(low[u],dfn[v]);
    }

    if(low[u] == dfn[u]){//找到极大联通分量
        scc++;
        while(top > 0&&stk[top] != u){
            top --;
            vis[stk[top] ] = 2;
            num[stk[top] ] = scc;
        }
    }

}
int main(){
    int n,m,i,j,k,a,b,c,cla;
    scanf("%d",&cla);
    for(int zu = 1;zu <=cla;++ zu){
        scanf("%d%d",&n,&m);
        init();
        for(i = 0;i <m;++ i){
            scanf("%d%d",&a,&b);
            Add(a,b);
        }
        for(i = 1;i <= n;++ i)
            if(!dfn[i])
                Tarjan(i);
        bool bj = false;
        for(i = 1;i <= n;++ i){
            for(j = head[i];~j; j = q[j].next){
                int v = q[j].to;
                if(num[i] != num[v]){
                    out[num[i] ]++;
                    in[num[v] ]++;
                }
            }

        }
        int ans1,ans2;
        ans1 = ans2 =0;
        for(i = 1;i <= scc;++ i){
            if(in[i]==0)
                ans1++;
            if(out[i]==0)
                ans2++;
            cout<<i<<endl;
        }
        printf("Case %d: ",zu);
        if(ans1>1||ans2>1){
            puts("The Burning Shadow consume us all");
        }
        else{
            puts("Kalimdor is just ahead");
        }
    }
    return 0;
}

SDUT 2604 Thrall’s Dream （单连通的判断）

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原文地址：http://blog.csdn.net/grit_icpc/article/details/51333543