HYSBZ_1854_并查集

Posted 2020-08-11 冷暖知不知

http://www.lydsy.com/JudgeOnline/problem.php?id=1854

 

每次判断每组两个数的根，若不等，则小的遍历1，大的为根，若相等，则说明前面的小的都遍历过，根遍历1。最后判断vis即可。

 

#include<iostream>
#include<cstdio>
using namespace std;
int pre[1000005],vis[1000005] = {0};

int findd(int x)
{
    int root = x;
    while(root != pre[root])    root = pre[root];
    int temp;
    while(x != pre[x])
    {
        temp = pre[x];
        pre[x] = root;
        x = temp;
    }
    return root;
}

void join(int x,int y)
{
    int xx = findd(x),yy = findd(y);
    if(xx == yy)    vis[xx] = 1;
    else
    {
        if(xx > yy) swap(xx,yy);
        pre[xx] = yy;
        vis[xx] = 1;
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= 100005;i++)   pre[i] = i;
    while(n--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        join(a,b);
    }
    int i;
    for(i = 1;i <= n;i++)
    {
        if(vis[i] == 0) break;
    }
    printf("%d",i-1);
    return 0;
}