带权并查集（裸

A very big corporation is developing its corporative network. In the beginning each of the N enterprises
of the corporation, numerated from 1 to N, organized its own computing and telecommunication center.
Soon, for amelioration of the services, the corporation started to collect some enterprises in clusters,
each of them served by a single computing and telecommunication center as follow. The corporation
chose one of the existing centers I (serving the cluster A) and one of the enterprises J in some other
cluster B (not necessarily the center) and link them with telecommunication line. The length of the
line between the enterprises I and J is |I −J|(mod 1000). In such a way the two old clusters are joined
in a new cluster, served by the center of the old cluster B. Unfortunately after each join the sum of the
lengths of the lines linking an enterprise to its serving center could be changed and the end users would
like to know what is the new length. Write a program to keep trace of the changes in the organization
of the network that is able in each moment to answer the questions of the users.
Your program has to be ready to solve more than one test case.
Input
The first line of the input file will contains only the number T of the test cases. Each test will start
with the number N of enterprises (5 ≤ N ≤ 20000). Then some number of lines (no more than 200000)
will follow with one of the commands:
E I — asking the length of the path from the enterprise I to its serving center in the moment;
I I J — informing that the serving center I is linked to the enterprise J.
The test case finishes with a line containing the word ‘O’. The ‘I’ commands are less than N.
Output
The output should contain as many lines as the number of ‘E’ commands in all test cases with a single
number each — the asked sum of length of lines connecting the corresponding enterprise with its serving
center.
Sample Input
1
4
E 3
I 3 1
E 3
I 1 2
E 3
I 2 4
E 3
O
Sample Output
0
2
3
5

题目大意
有n个节点，然后执行I u，v（把u的父节点设为v）和E u（询问u到根节点的距离）
共有两种操作

解题思路
路径压缩时顺便更新距离就可以 但是一定要先回溯，之前把回溯放后面了 样例过了但是一直wa

加粗样式

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e5;
int f[maxn];
int d[maxn];
int find(int x)
{
	if(x!=f[x])
	{
		int t=f[x];
		f[x]=find(f[x]);
		d[x]+=d[t];
	}
	return f[x];
} 
int main()
{
	int x,y,n,t,i;
	char a[5];
	cin>>t;
	while(t--)
	{
		cin>>n;
		for(i=1;i<=n;i++)
		{
			f[i]=i;
			d[i]=0;
		}
		while(cin>>a)
		{
			if(a[0]=='O')
			 break;
			if(a[0]=='I')
			{
				cin>>x>>y;
				f[x]=y;
				d[x]=abs(x-y)%1000;
			}
			if(a[0]=='E')
			{
				cin>>x;
				find(x);
				cout<<d[x]<<endl;
			}
		}
	}
	return 0;
}