HDU 2689 HDOJ 2689 Sort it ACM 2689 IN HDU

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题目地址 :

      http://acm.hdu.edu.cn/showproblem.php?pid=2689 

题目描述:

   其实就是求 冒泡排序时 的交换次数,  当然也可以求逆序数来解决问题, 下面是2份 代码:

 

//直接冒泡排序求交换的次数
/*
Mail to   : miyubai@gamil.com
My Blog   : www.baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_2689
Doc Name  : Sort it
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
int N, num[1010];
inline void swap ( int &a, int &b ) {
       a ^= b ^= a ^= b;       
}
int bouble () {
    int sum = 0;
    for ( int i = 0; i < N; ++ i ) {
         for ( int j = 1; j < N - i; ++ j ) {
              if ( num[j-1] > num[j] ) {
                   swap ( num[j-1], num[j] );
                   ++ sum;
              }    
         }    
    }    
    return sum;
}
void print () {
     for ( int i = 0; i < N; ++ i )
     cout << num[i] << " ";
     cout << endl;     
}
int main ()
{
    while ( scanf ( "%d", &N ) == 1 ) {
           for ( int i = 0; i < N; ++ i ) {
                scanf ( "%d", num + i );    
           }       
           printf ( "%d\n",bouble () );
          // print ();
    }
    return 0;
}

//树状数组求逆序数法
/*
Mail to   : miyubai@gamil.com
My Blog   : www.baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_2689
Doc Name  : Sort it
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
int N,val,num[1010],low[1010];
void init () {
     for ( int i = 0; i <= 1010; ++ i ) {
          low[i] = i & ( -i );    
     }
}
void modify ( int x ) {
     while ( x <= N ) {
            ++ num[x];      
            x += low[x];
     }     
}
int query ( int x ) {
    int sum = 0;
    while ( x > 0 ) {
           sum += num[x];
           x -= low[x];      
    }    
    return sum;
}
int main ()
{
    init ();
    while ( scanf ( "%d", &N ) == 1 ) {
           memset ( num, 0, sizeof ( num ) );  
           int sum = 0;
           for ( int i = 0; i < N; ++ i ) {
                scanf ( "%d", &val );
                modify ( val ); 
                sum += i - query ( val - 1 );   
           }  
           printf ( "%d\n", sum );
    }
     
    return 0;
}