• 3027


        

    3027 - Corporative Network

      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

    题意:没有更水,只有最水。。。

    就是进行一系列操作,I把u的父节点设为v,权值是(v-u)%1000;

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    #define mem(x,y) memset(x,y,sizeof(x))
    #define SI(x) scanf("%d",&x)
    const int MAXN=20010;
    int pre[MAXN];
    int ans;
    /*int find(int x){
    	return x==pre[x]?x:find(pre[x]);
    }*/
    void merge(int x,int y){
    	pre[x]=y;
    }
    int main(){
    	int T,N,a,b;
    	SI(T);
    	char s[2];
    	while(T--){
    		mem(pre,0);
    		SI(N);
    		while(scanf("%s",s),s[0]!='O'){
    			if(s[0]=='I'){
    				SI(a);SI(b);
    				merge(a,b);
    			}
    			else if(s[0]=='E'){
    				SI(a);ans=0;
    				while(a!=pre[a]&&pre[a]!=0){
    					ans+=(abs(a-pre[a])%1000);
    					a=pre[a];
    				}
    				printf("%d
    ",ans);
    			}	
    		}
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    软件架构学习小结
    20+ 个很有用的 jQuery 的 Google 地图插件 (英语)
    网页JS获取当前地理位置(省市区)
    前端Js框架汇总(工具多看)
    MUI简介-最接近原生App体验的前端框架
    Bootstrap手机网站开发案例
    jQuery Mobile手机网站案例
    历届图灵奖 (Turing award)得奖名单
    js进阶 10-9 -of-type型子元素伪类选择器
    网页如何实现隔多久自动调用某个方法
  • 原文地址:https://www.cnblogs.com/handsomecui/p/5020108.html
Copyright © 2020-2023  润新知