题意:有n个节点,初始话每个节点的父节点都是不存在的,你的任务是执行I或者E操作
I:u,v将u的父节点设为v ,距离为|u-v|%1000; E:询问u到根节点的距离
输出每条E操作
思路:在并查集的基础上加上路径的压缩
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int mod = 1000;
const int MAXN = 20005;
int d[MAXN],f[MAXN],n;
int find(int x){
if (f[x] != x){
int root = find(f[x]);
d[x] += d[f[x]];
return f[x] = root;
}
else return x;
}
int main(){
int t;
scanf("%d",&t);
while (t--){
scanf("%d",&n);
for (int i = 0; i <= n; i++)
f[i] = i,d[i] = 0;
int u,v;
char op;
while (cin >> op && op != 'O'){
if (op == 'I'){
scanf("%d%d",&u,&v);
f[u] = v;
d[u] = abs(u-v) % mod;
}
else {
scanf("%d",&u);
find(u);
printf("%d\n",d[u]);
}
}
}
return 0;
}