hdu1269 迷宫城堡 (强连通分量Tarjan算法)

来源：互联网发布：linux 重启 reboot 编辑：程序博客网时间：2024/04/28 16:06

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

题解：对于任意的i和j，至少存在一条路径可以从房间i到房间j，也存在一条路径可以从房间j到房间i。求强连通分量==1？

#include <stdio.h>#include <string.h>#define MAXN 10001struct node{int from,to,next;}edge[10*MAXN];int head[MAXN],instack[MAXN],low[MAXN],dfn[MAXN];int stack[MAXN],tot,Dindex,top,Bcnt;void Init(){//初始化tot=0,top=0,Dindex=0,Bcnt=0;memset(head,-1,sizeof(head));memset(instack,0,sizeof(instack));memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));}void addEdge(int from,int to){edge[tot].from=from;edge[tot].to=to;edge[tot].next=head[from];head[from]=tot++;}void Tarjan(int x){int i,u,temp;dfn[x]=low[x]=++Dindex;//时间戳stack[top++]=x;instack[x]=1;for(i=head[x];i!=-1;i=edge[i].next){u=edge[i].to;if(!dfn[u]){Tarjan(u);low[x]=low[x]>low[u]?low[u]:low[x];}else if(instack[u]&&low[x]>dfn[u])low[x]=dfn[u];}if(low[x]==dfn[x]){Bcnt++;do {temp=stack[--top];instack[temp]=0;} while (x!=temp);}}int Scan()        {        char ch;        int ret=0;        while((ch=getchar())<'0'||ch>'9');        while(ch>='0'&&ch<='9')        {        ret=ret*10+(ch-'0');        ch=getchar();        }        return ret;        }   int main(){int n,m,i,x,y;while(scanf("%d %d",&n,&m)&&(n+m)!=0){Init();while(m--){//scanf("%d %d",&x,&y);x=Scan();y=Scan();addEdge(x,y);}for(i=1;i<=n;++i){if(!dfn[i])Tarjan(i);}if(Bcnt==1)printf("Yes\n");elseprintf("No\n");}return 0;}