HDU 迷宫城堡 1269 （强连通图判定）

来源：互联网发布：linux mv 创建目录编辑：程序博客网时间：2024/04/30 17:39

大意：问图中是不是所有两点都是任意可达的，i到j,同样j也要到达i.

就是缩成一个点即可，所以只需要判断scc是不是1即可。不需要找麻烦看度数。

#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|1using namespace std;const int Ma = 10100;struct node{    int to,w,next;}q[Ma*15];int head[Ma*15],dfn[Ma],num[Ma],du[Ma],stk[Ma],vis[Ma],low[Ma];int cnt,top,tim,scc,out[Ma];bool bj;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;bj = false;    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(num,0,sizeof(num));    memset(out,0,sizeof(out));    memset(vis,0,sizeof(vis));    memset(low,0,sizeof(low));}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]){        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,cla;    while(~scanf("%d%d",&n,&m)){        if(!n&&!m) break;        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);        if(scc==1)            puts("Yes");        else            puts("No");    }    return 0;}

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