题目大意：
查询某1个节点到根节点之间的距离
解题思路：
加权并查集问题。之前做的题目是“查看两个或多个节点是不是在同1个集合下”，现在的题目是“查询某个节点到
根节点之间的距离”。之前只需要使用到father[x]这个数组，用来表示x的父亲节点是谁。现在引入dist[x]数组，用来记录
x节点到根节点的距离
1）在并查集中，根节点不懂，其他节点都可以动。
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 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.
Input
Your program has to be ready to solve more than one test case. 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
代码以下：
* LA3027.cpp
*
* Created on: 2015年1月3日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 20005;
int father[maxn];
int dist[maxn];
int find(int x) {
if (x != father[x]) {
int fx = find(father[x]);
dist[x] += dist[father[x]];
father[x] = fx;
}
return father[x];
}
int main() {
int t;
scanf("%d", &t);
while (t–) {
char cmd[9];
int n;
scanf("%d", &n);
int i;
for (i = 1; i <= n; ++i) {
father[i] = i;
dist[i] = 0;
}
while (scanf("%s", cmd) == 1, cmd[0] != ‘O’) {
if (cmd[0] == ‘I’) {
int u, v;
scanf("%d%d", &u, &v);
father[u] = v;
dist[u] = abs(u – v) % 1000;
} else if (cmd[0] == ‘E’) {
int u;
scanf("%d", &u);
find(u);
printf("%d
", dist[u]);
}
}
}
return 0;
}
波比源码 » (DS 《算法竞赛入门经典》)LA 3027 Corporative Network(查询某一个节点到根节点之间的距离)
