04-树5. File Transfer (25)

题目来源：

We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?

Input Specification:

Each input file contains one test case. For each test case, the first line contains N (2<=N<=10⁴), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format:

I c1 c2  

where I stands for inputting a connection between c1 and c2; or

C c1 c2    

where C stands for checking if it is possible to transfer files between c1 and c2; or

S

where S stands for stopping this case.

Output Specification:

For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network.

Sample Input 1:

5C 3 2I 3 2C 1 5I 4 5I 2 4C 3 5S

Sample Output 1:

nonoyesThere are 2 components.

Sample Input 2:

5C 3 2I 3 2C 1 5I 4 5I 2 4C 3 5I 1 3C 1 5S

Sample Output 2:

nonoyesyesThe network is connected.

java解：  几个测试点会 超时

import java.util.Arrays;import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;public class Main{public static void main(String[] args){Scanner scanner = new Scanner(System.in);//主机数int n = scanner.nextInt();//节点i的父节点 father[i]int[] father = new int[n+1];for(int i=1;i<=n;i++){father[i]=i;}//操作String s ;//再输入S时停止while(  !(s=scanner.next()).equals("S") ){int a = scanner.nextInt();int b = scanner.nextInt();//如果输入的是Cif(s.equals("C")){//查找a,b的根（最上层父元素）if( findRoot(a,father)==findRoot(b,father)){System.out.println("yes");}else{System.out.println("no");}}//如果输入的是Iif(s.equals("I")){//如果a,b不同根if( !(findRoot(a,father)==findRoot(b,father))){father[a]=b;n--;}}}if(n==1){System.out.println("The network is connected.");}else{System.out.println("There are "+n+" components.");} }public static int findRoot(int a ,int[] father){while(!( a==father[a])){a= father[a];}return father[a];}}  

c++ 解：

#include <bits/stdc++.h>using namespace std;const int maxn = 10005;int father[maxn];//int findroot(int x) {//    if(x != father[x]) {//        x = findroot(father[x]);//    }//    return x;//}int findroot(int x) {    if(x != father[x]) {        father[x] = findroot(father[x]);    }    return father[x];}int main() {    int n, a, b;    char s[2];    scanf("%d", &n);    for (int i = 1; i <= n; i ++) {        father[i] = i;    }    while (scanf("%s", s) != EOF && s[0] != 'S') {        scanf("%d%d", &a, &b);        if(s[0] == 'C') {            if(findroot(a) != findroot(b)) {                printf("no\n");            } else {                printf("yes\n");            }        }        if(s[0] == 'I') {            int ar = findroot(a);            int br = findroot(b);            if(ar != br) {                father[ar] = br;                n --;            }        }    }    if(n == 1) {        printf("The network is connected.\n");    } else {        printf("There are %d components.\n", n);    }    return 0;}

http://blog.csdn.net/u012762625/article/details/43817447