HDU1325 &&poj1308 基础并查集

来源：互联网发布：翻译硕士学校排名知乎编辑：程序博客网时间：2024/05/03 23:09

和上一道小希的迷宫差不多，但是在HDU上提交一直WA，POJ过了

HDU的数据太强了吧,强的我感觉数据有问题

题意：输入若干对点，判断是否是一颗树，转化过来也就是是否存在环

点数-边数=1，还要判断每个点的入度都<=1

POJ AC代码

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <algorithm>const int INF = 1e8;using namespace std;int father[1010];bool vis[1010];int rudu[1010];int findx(int r){    int i = r,j;    while(father[r]!=r)    {         r=father[r];    }    while(father[i]!=r)    {        j = father[i];        father[i] = r;        i = j;    }    return r;}bool Merge(int x,int y){    int fx,fy;    fx=findx(x);    fy=findx(y);    if(fx!=fy){ father[fx]=fy; return 1;}elsereturn 0;}void init(){    for(int i=0;i<1010;i++){father[i]=i;vis[i] = 0 ;rudu[i] = 0;}}int main(){int a,b,c = 0;while(scanf("%d%d",&a,&b)!=EOF){    c++;    if(a<0&&b<0)            break;    intflag=1,t=0;         if(a==0 && b==0)        {            printf("Case %d ",c);            puts("is a tree.");            continue;        }init();    intnum = 0;while(1){    if(a==0&&b==0) break;    if(flag==1)            {if(!vis[a]) {num++; } if(!vis[b]) {num++; rudu[b]++;}if(rudu[b]>1)                flag = 0;vis[a]=1; vis[b]=1;           if(Merge(a,b)==1)           {                t++;               }            else                flag = 0;            }   scanf("%d%d",&a,&b);}printf("Case %d ",c);if(num-t==1 &&flag == 1)puts("is a tree.");elseputs("is not a tree.");}return 0;}

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