PAT 04-2. File Transfer (25) （并查集）

04-2. File Transfer (25)

题目链接：http://www.patest.cn/contests/mooc-ds/04-2

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 andc2; or

C c1 c2

where C stands for checking if it is possible to transfer files betweenc1 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 betweenc1 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." wherek is the number of connected components in this network.

Sample Input 1:

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

Sample Output 1:

no
no
yes
There are 2 components.

Sample Input 2:

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

Sample Output 2:

no
no
yes
yes
The network is connected.

解题报告：

裸并查集，不过好久不写，路径压缩写漏了，一个点错了一次。。。。

#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;
}