程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ1962:Corporative Network(並查集)

POJ1962:Corporative Network(並查集)

編輯：C++入門知識

POJ1962:Corporative Network(並查集)

Description

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.

Input

Your program has to be ready to solve more than one test case. The first line of the input 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

Source

Southeastern Europe 2004
題意:有n個點，一開始每個點以自己為信號終點，當輸入I X Y，就連接X,Y，並且X的信號終點指向Y，E X是輸出X距離信號終點的距離
思路:經典並查集，只是每個節點帶權，需要更新

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define Len 20005
#define mod 19999997
const int INF = 0x3f3f3f3f;

int t,n;
char str[5];
int father[Len],dis[Len];

int find(int x)
{
    if(x == father[x])
        return x;
    int fx = find(father[x]);
    dis[x] = dis[x]+dis[father[x]];
    return father[x]=fx;

}

void solve(int x,int y)
{
    int fx = find(x),fy = find(y);
    if(fx == fy)
    return;
    father[x] = x;
    dis[fx] = dis[x];
    father[x] = y;
    dis[x] = abs(x-y)%1000;
}

int main()
{
    int i,j,k,x,y;
    scanf("%d",&t);
    w(t--)
    {
        scanf("%d",&n);
        up(i,0,n)
        {
            father[i] = i;
            dis[i] = 0;
        }
        w(~scanf("%s",str))
        {
            if(str[0]=='O')
                break;
            if(str[0]=='E')
            {
                scanf("%d",&x);
                find(x);
                printf("%d\n",dis[x]);
            }
            else
            {
                scanf("%d%d",&x,&y);
                solve(x,y);
            }
        }
    }

    return 0;
}