并查集

SZU Problem(A70):Locked Boxes

Judge Info

Memory Limit: 32768KB
Case Time Limit: 10000MS
Time Limit: 10000MS
Judger: Number Only Judger

Description

Poseidon has N boxes with lock. What's inside the box? (hum..., think by yourself) Each box can either be opened with its corresponding key or smashed. Poseidon has put the keys in some of the boxes. He remembers which key has been placed in which box. Poseidon wants to access to all of the boxes. However, he wants to destroy as few of them as possible. Please write a Program to help Poseidon to determine how many boxes have to be smashed.

Input

The input will contains multiple test cases. The first line of the input is a single integer $T (1 leq T leq 30)$ which is the number of test cases. T test cases follow.

Each test case contains a single integer $N (1 leq N leq 1,000,000)$ - this is the number of boxes owned by Poseidon. The boxes (as well as their corresponding keys) are numbered from 1 to N. Next, there are N lines: the i+1st line contains a single integer - the number of the box which the ith key has been placed in.

Output

For each input test case, you are to output a single integer - the minimal number of boxes to be smashed in order to access to all of the boxes.

Sample Input

Sample Output

2
2

 1 #include <stdio.h>
 2 #define N 1000010
 3 int array[N];  // 用来记录箱子的钥匙在哪里
 4 int search( int );
 5  
 6 int main()
 7 {
 8     int t, n, i, a;
 9     int num = 0; // 计算需要打烂多少个箱子
10     scanf( "%d", &t );
11     while( t-- )
12     {
13         scanf( "%d", &n );
14         // 初始化
15         for( i=1; i<=n; i++ )
16             array[i] = i;
17         num = 0;
18         // 输入箱子的钥匙在哪里并判断它的根是谁。。
19         for( i=1; i<=n; i++ )
20         {
21             scanf( "%d", &a );
22             array[i] = search(a);
23             if( array[i] == i )
24                 num++;
25         }
26         printf( "%d
", num );
27     }
28     return 0;
29 }
30  
31 int search( int i )
32 {
33     int key = i;
34     // 找根
35     while( array[i] != i )
36         i = array[i];
37     // 路径压缩
38     int root = i;
39     int t = key;
40     int temp;
41     while( array[t] != t )
42     {
43         temp = array[t];
44         array[t] = root;
45         t = temp;
46     }
47     return root;
48 }

SZU Problem(B36):Reading books

Judge Info

Memory Limit: 32768KB
Case Time Limit: 10000MS
Time Limit: 10000MS
Judger: Number Only Judger

Description

In the summer vacation, LRJ wants to improve himself in computer science. So he finds out N books of computer science in the school library. The books are numbered from 0 to N-1.

To finish reading the i-th book, it takes LRJ time[i] minutes. But some books are similar in the content. If the i-th book and the j-th book are similar, then if LRJ has finished reading the i-th book, it will take him only $left lfloor frac{time[j]}{2} ight floor$ minutes to finish reading the j-th book. Of course if LRJ has finished reading the j-th book, it will take him only $left lfloor frac{time[i]}{2} ight floor$ minutes to finish reading the i-th book. Now you are asked to tell LRJ the minimal total time to finish reading all the N books.

Input

The first line contains two integers $N(0leq N leq 100)$ and $M(0leq M leq N imes (N-1)/2)$ . N is the total number of books. M is the number of pairs which are similar.

Then the following N lines describe $time[0],time[1],cdots,time[n-1](1leq time[i] leq 10^{5})$ .

Next comes M lines, each contains two integer (i,j), indicating that the i-th book and the j-th book are similar.

Input is ended with EOF.

Output

For each test case, just output the minimal total time on a single line.

Sample Input

Sample Output

11
3
10

 1 #include <stdio.h>
 2 #define N 110
 3 #define M 5000
 4 int time[N];
 5 int array[M];
 6 int sum[N]; // 按秩分配
 7 int search( int );
 8 int main()
 9 {
10     int n, m;
11     int i;
12     int a, b;
13     int total;
14     while( scanf("%d%d", &n, &m ) != EOF )
15     {
16         total = 0;
17         for( i=0; i<n; i++ )
18         {
19             scanf( "%d", &a );
20             time[i] = a;
21             sum[i] = time[i]/2;
22         }
23         for( i=0; i<n; i++ )
24             array[i] = i;
25         i = 0;
26         while( i < m )
27         {
28             scanf( "%d%d", &a, &b );
29             // 记得比较祖先~~~
30             a = search(a);
31             b = search(b);
32             i++;
33             // 记得考虑两者拥有共同祖先的时候!!!
34             if( a == b )
35                 continue;
36             if( time[a] <= time[b] )
37             {
38                 array[b] = a;
39                 sum[a] += sum[b];
40             }
41             else
42             {
43                 array[a] = b;
44                 sum[b] += sum[a];
45             }
46         }
47         for( i=0; i<n; i++ )
48             if( array[i] == i )
49             {
50                 sum[i] = sum[i] - time[i]/2 + time[i];
51                 total += sum[i];
52             }
53         printf( "%d
", total );
54     }
55     return 0;
56 }
57  
58 int search( int i )
59 {
60     int t = i;
61     // 找根
62     while( array[i] != i )
63         i = array[i];
64     // 路径压缩
65     int root = i;
66     int temp;
67     while( array[t] != t )
68     {
69         temp = array[t];
70         array[t] = root;
71         t = temp;
72     }
73     return root;
74 }

SZU Problem(B64):Network Connections

Judge Info

Memory Limit: 32768KB
Case Time Limit: 10000MS
Time Limit: 10000MS
Judger: Normal

Description

Bob, who is a network administrator, supervises a network of computers. He is keeping a log connections between the computers in the network. Each connection is bi-directional. Two computers are interconnected if they are directly connected or if they are interconnected with the same computer. Occasionally, Bob has to decide, quickly, whether two given computers are connected, directly or indirectly, according to the log information.

Write a program which based on information input from a text file counts the number of successful and the number of unsuccessful answers to the questions of the kind : is computer $i \,$ interconnected with computer $j\,$ ?

Input

The first line of the input contains the number of dataset, and it's followed by a blank line. Each dataset is defined as follows:

The number of computers in the network $N\,$ ( $N leq 10,000$ );
A list of pairs of the form:
- $c\,$ computer $i\,$ computer $j\,$ , where computer $i\,$ and computer $j\,$ are integers from $1\,$ to $N\,$ . A pair of this form shows that computer $i\,$ and computer $j\,$ get interconnected.
- $q\,$ computer $i\,$ computer $j\,$ , where computer $i\,$ and computer $j\,$ are integers from $1\,$ to $N \,$ . A pair of this form stands for the question: is computer $i\,$ interconnected with computer $j\,$ ?

There's a blank line between datasets.

Each pair is on a separate line. Pairs can appear in any order, regardless of their type. The log is updated after each pair of type (a) and each pair of type (b) is processed according to the current network configuration.

Output

For example, the input file illustrated in the sample below corresponds to a network of 10 computers and 7 pairs. There are $N1\,$ successfully answered questions and $N2\,$ unsuccessfully answered questions. The program prints these two numbers to the standard output on the same line, in the order: successful answers, unsuccessful answers, as shown in the sample output. Print a blank line between datasets.

Sample Input

Sample Output

1,2
血的教训，能用getchar(), 决不用scanf( "%c", &c );

 1 #include <stdio.h>
 2 #define N 10005
 3 int array[N];
 4 int findroot( int );
 5  
 6 int main()
 7 {
 8     int t, a, b, i, n, yes, no;
 9     char request;
10     scanf( "%d", &t );
11     while( t-- )
12     {
13         scanf( "%d
", &n );
14         // 初始化啊，我是石头里蹦出来的，自己爸爸的是自己^^
15         for( i=1; i<=n; i++ )
16             array[i] = i;
17         yes = no = 0;
18         while( request = getchar() )
19         {
20             // 手拉手我们是好朋友
21             if( request == 'c' )
22             {
23                 scanf( "%d%d", &a, &b );
24                 if( findroot(a) != findroot(b) ) // 判断是否同祖先
25                     array[findroot(a)] = findroot(b);
26             }
27             // 问呀问呀问问题
28             else if( request == 'q' )
29             {
30                 scanf( "%d%d", &a, &b );
31                 if( a == b )
32                     yes++;
33                 else
34                 {
35                     if( findroot(a) == findroot(b) )
36                         yes++;
37                     else
38                         no++;
39                 }
40             }
41             else  // 最后一个不判断
42                 break;
43             getchar();
44         }
45         printf( "%d,%d
", yes, no );
46         if( t )
47             printf( "
" );
48     }
49     return 0;
50 }

SZU Problem(J21):Are they the same?

Judge Info

Memory Limit: 65536KB
Case Time Limit: 10000MS
Time Limit: 10000MS
Judger: Normal

Description

You are given N relations about two words. Here relation means the two words can change from each other. Then I have m times queries about whether given two strings are the same. Every string has no more than 100 words. Every word has no more than 10 letters. To make two strings same, you can exchange the position of any two words in the string.

Input

The number of the relations N (1<=N<=10000), and what they are.

The number of the test case M (1<=M<=10000) , and what they are. Please read sample for more.

Output

For each case out put “yes” if the two string may be same, otherwise, output “no”. You should also output “Case + one space + No. + : + one space” before the answer.

For more details, please read the sample.

Sample Input

3
i wo
love ai
you ni
5
ni
wo
i love you
wo ai ni
i love you
ni ai wo ma
i love you
ni ai wo
i love you
ni wo

Sample Output

Case 1: no
Case 2: yes
Case 3: no
Case 4: yes
Case 5: no

#include <iostream>
#include <map>
#include <string>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 10005;
int array[N];
int buffer[N];
int root[N];
int findroot( int );
int main()
{
    int n, i, j, m, p, q, num = 0;
    string temp1, temp2;
    map<string, int> magic;
    while( scanf( "%d", &n ) != EOF )
    {
        // 我的爸爸就是我^^
        for( i=1; i<=n; ++i )
            root[i] = i;
 
        for( i=1; i<=n; ++i )
        {
            cin >> temp1;
            if( magic[temp1] == 0 )
                magic[temp1] = i;
            cin >> temp2;
            if( magic[temp2] == 0 )
                magic[temp2] = i;
            p = findroot(magic[temp1]);
            q = findroot(magic[temp2]);
            if( p != q )
                root[p] = q;
        }
        scanf( "%d", &m );
        for( i=1; i<=m; ++i )
        {
            p = q = 0;
            while( cin >> temp1 )
            {
                array[p++] = root[magic[temp1]];
                if( getchar() == '
' )
                    break;
            }
            while( cin >> temp1 )
            {
                buffer[q++] = root[magic[temp1]];
                if( getchar() == '
' )
                    break;
            }
 
            if( p != q )
                cout << "Case " << ++num << ": no" << endl;
            else
            {
                sort( array, array+p );
                sort( buffer, buffer+q );
                for( j=0; j<p; ++j )
                    if( array[j] != buffer[j] )
                        break;
 
                if( j == p )
                    cout << "Case " << ++num << ": yes" << endl;
                else
                    cout << "Case " << ++num << ": no" << endl;
            }
        }
    }
    return 0;
}
int findroot( int i )
{
    int n = i;
    while( root[n] != n )
        n = root[n];
    int t = i;
    int temp;
    while( root[t] != t )
    {
        temp = root[t];
        root[t] = n;
        t = temp;
    }
    return n;
}

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原文地址：https://www.cnblogs.com/zhongshuxin/p/3276773.html