「SP25784」BUBBLESORT - Bubble Sort 解题报告

SP25784 BUBBLESORT - Bubble Sort

题目描述

One of the simplest sorting algorithms, the Bubble Sort, can be expressed as (0-based array):

procedure bubbleSort( A : list of sortable items )

n = length(A)

repeat

swapped = false

for i = 1 to n-1 inclusive do

/* if this pair is out of order */

if A[i-1] > A[i] then

/* swap them and remember something changed */

swap( A[i-1], A[i] )

swapped = true

end if

end for

until not swapped

end procedure

Now, given an array of N integers, you have to find out how many swap opeartions occur if the Bubble Sort algorithm is used to sort the array.

输入输出格式

输入格式：

Input begins with a line containing an integer T(1<=T<=100), denoting the number of test cases. Then T test cases follow. Each test case begins with a line containing an integer N(1<=N<=10000), denoting the number of integers in the array, followed by a line containing N space separated 32-bit integers.

输出格式：

For each test case, output a single line in the format Case X: Y, where X denotes the test case number and Y denotes the number of swap operations needed modulo 10000007.

输入输出样例

输入样例#1：

1
4
3 2 1 4

输出样例#1：

Case 1: 3

---

思路

运用归并排序，求出逆序对即可。具体看代码注释。

代码

#include<bits/stdc++.h>
using namespace std;
#define mod 10000007
#define MAXN 10005

int T, N, ans;
int a[MAXN], b[MAXN];

void Merge_sort( int l, int r ){
	if ( l == r ) return;//只有一个元素不用排序
	int mid((l + r) >> 1);
	Merge_sort( l, mid );//左边排序
	Merge_sort( mid + 1, r );//右边排序
	int i1(l), i2(mid + 1), c(l);//把左右两边归并
	while( i1 <= mid && i2 <= r ){
		if ( a[i1] <= a[i2] ) b[c++] = a[i1++];//必须写<=
		else b[c++] = a[i2++], ans = ( ans + mid - i1 + 1 ) % mod;//顺便统计逆序对个数（右边元素之前有mid-i1+1个比它大的）
	}
	while( i1 <= mid ) b[c++] = a[i1++];//剩余的元素排到后面
	while( i2 <= r ) b[c++] = a[i2++];
	for ( int i = l; i <= r; ++i ) a[i] = b[i], b[i] = 0;//复制回原数组
}

int main(){
	scanf( "%d", &T );
	for ( int Ti = 1; Ti <= T; ++Ti ){
		scanf( "%d", &N );
		for ( int i = 1; i <= N; ++i ) scanf( "%d", &a[i] );
		ans = 0;//初始化很重要
		Merge_sort( 1, N );
		printf( "Case %d: %d\n", Ti, ans );
	}
	return 0;
}