LA 3027 Corporative Network 合作网络【并查集+路径压缩】

LA 3027 Corporative Network 合作网络【并查集+路径压缩】http://www.bnuoj.com/bnuoj/problem_show.php?pid=9203

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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

14E 3I 3 1E 3I 1 2E 3I 2 4E 3O

Sample Output

0235

Source

Regionals 2004, Europe - Southeastern

【题意】有n个点，初始时每个节点的父节点都不存在，接下来有一系列操作：E操作和 I 操作；I u v：把节点u的父节点设为v，距离为|u-v|%1000，输入保证执行命令前u没有父节点； E u：询问u到根节点的距离。指令操作以O结束。

【思路】用并查集，记录附加信息；记下每个结点到父结点的距离为d【i】，然后在路径压缩时维护数组距离d

【代码如下】

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int maxn = 20000+10;int pa[maxn], d[maxn];int findset(int x){//路径压缩，同时维护d[i]    if(pa[x]!=x){        int root = findset( pa[x] );        d[x] += d[ pa[x] ];        return pa[x] = root;    }    else return x;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        int n, u, v;        char cmd[9];        scanf("%d", &n);        //初始化，每个结点单独是棵树        for(int i=1; i<=n; i++){            pa[i] = i; d[i] = 0;        }        while(scanf("%s", &cmd) && cmd[0]!='O'){            if(cmd[0]=='E'){                scanf("%d", &u);                findset(u);                printf("%d\n", d[u]);            }            if(cmd[0]=='I'){                scanf("%d %d", &u, &v);                pa[u] = v;                d[u] = abs(u-v)%1000;            }        }    }    return 0;}