传送门

Description

Background 
Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just by glancing at them. And he can even count Poker cards. Charlie would love to be able to do cool things like that, too. He wants to beat his brother in a similar task.

Problem 
Here's what Charlie thinks of. Imagine you get a sequence of N numbers. The goal is to move the numbers around so that at the end the sequence is ordered. The only operation allowed is to swap two adjacent numbers. Let us try an example:

Start with: 2 8 0 3 
swap (2 8) 8 2 0 3 
swap (2 0) 8 0 2 3 
swap (2 3) 8 0 3 2 
swap (8 0) 0 8 3 2 
swap (8 3) 0 3 8 2 
swap (8 2) 0 3 2 8 
swap (3 2) 0 2 3 8 
swap (3 8) 0 2 8 3 
swap (8 3) 0 2 3 8

So the sequence (2 8 0 3) can be sorted with nine swaps of adjacent numbers. However, it is even possible to sort it with three such swaps:

Start with: 2 8 0 3 
swap (8 0) 2 0 8 3 
swap (2 0) 0 2 8 3 
swap (8 3) 0 2 3 8

The question is: What is the minimum number of swaps of adjacent numbers to sort a given sequence?Since Charlie does not have Raymond's mental capabilities, he decides to cheat. Here is where you come into play. He asks you to write a computer program for him that answers the question. Rest assured he will pay a very good prize for it.

Input

The first line contains the number of scenarios. 
For every scenario, you are given a line containing first the length N (1 <= N <= 1000) of the sequence,followed by the N elements of the sequence (each element is an integer in [-1000000, 1000000]). All numbers in this line are separated by single blanks.

Output

Start the output for every scenario with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the minimal number of swaps of adjacent numbers that are necessary to sort the given sequence. Terminate the output for the scenario with a blank line.

Sample Input

4
4 2 8 0 3
10 0 1 2 3 4 5 6 7 8 9
6 -42 23 6 28 -100 65537
5 0 0 0 0 0

Sample Output

Scenario #1:
3

Scenario #2:
0

Scenario #3:
5

Scenario #4:
0

思路

最小相邻交换次数=逆序数

 
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const int maxn = 1005;
int sum = 0;
void merge_array(int array[],int left,int mid,int right)
{
	if (left >= right)	return;
	int i = left,j = mid + 1,k = 0;
	int *p;
	p = (int *)malloc((right-left+1)*sizeof(int));
	while (i <= mid && j <= right)
	{
		if (array[i] <= array[j])	p[k++] = array[i++];
		else	p[k++] = array[j++],sum += mid - i + 1;       //[i,mid]都能与array[j]形成逆序数
	}
	while (i <= mid)	p[k++] = array[i++];
	while (j <= right)	p[k++] = array[j++];
	for (i = 0;i < k;i++)	array[i+left] = p[i];
	free(p);
}

void merge_sort(int array[],int left,int right)
{
	if (left >= right)	return;
	int mid = left + ((right - left)>>1);
	merge_sort(array,left,mid);
	merge_sort(array,mid+1,right);
	merge_array(array,left,mid,right);
}

int main()
{
	int T,Case = 0;
	scanf("%d",&T);
	while (T--)
	{
		int N,a[maxn];
		scanf("%d",&N);
		for (int i = 0;i < N;i++)	scanf("%d",&a[i]);
		sum = 0;
		merge_sort(a,0,N-1);
		printf("Scenario #%d:\n%d\n\n",++Case,sum);
	}
	return 0;
}

  

POJ 1804 Brainman(归并排序)的更多相关文章

  1. POJ 1804 Brainman(5种解法,好题,【暴力】,【归并排序】,【线段树单点更新】,【树状数组】,【平衡树】)

    Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10575   Accepted: 5489 Descrip ...

  2. POJ 1804 Brainman

    Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7787   Accepted: 4247 Descript ...

  3. poj 1084 Brainman(归并排序)

    题目链接:http://poj.org/problem?id=1804 思路分析:序列的逆序数即为交换次数,所以求出该序列的逆序数即可. 根据分治法思想,序列分为两个大小相等的两部分,分别求子序列的逆 ...

  4. poj 1804 (nyoj 117)Brainman : 归并排序求逆序数

    点击打开链接 Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7810   Accepted: 4261 D ...

  5. POJ 1804 逆序对数量 / 归并排序

    Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 12175   Accepted: 6147 Descrip ...

  6. 【POJ 1804】 Brainman

    [题目链接] 点击打开链接 [算法] 本题是一个很经典的问题 : 归并排序求逆序对数,可以用分治算法解决 分治,分而治之,分治算法的思想就是将一个问题转化为若干个子问题,对这些子问题分别求解,最后, ...

  7. POJ 1840 Brainman(逆序对数)

    题目链接:http://poj.org/problem?id=1804 题意:给定一个序列a[],每次只允许交换相邻两个数,最少要交换多少次才能把它变成非递降序列. 思路:题目就是要求逆序对数,我们知 ...

  8. POJ 2299 Ultra-QuickSort 归并排序、二叉排序树,求逆序数

    题目链接: http://poj.org/problem?id=2299 题意就是求冒泡排序的交换次数,显然直接冒泡会超时,所以需要高效的方法求逆序数. 利用归并排序求解,内存和耗时都比较少, 但是有 ...

  9. poj 2299 Ultra-QuickSort 归并排序求逆序数对

    题目链接: http://poj.org/problem?id=2299 题目描述: 给一个有n(n<=500000)个数的杂乱序列,问:如果用冒泡排序,把这n个数排成升序,需要交换几次? 解题 ...

随机推荐

  1. Burndown chart

    S型的燃尽图 在一次milestone开发过程中,开发者会持续编辑issue列表,每个issue都有自己的生命周期.燃尽图预期这些issues会被线性的消灭掉,所以从第一天直接到最后一天画个直线表示预 ...

  2. node 学习笔记 - Modules 模块加载系统 (2)

    本文同步自我的个人博客:http://www.52cik.com/2015/12/14/learn-node-modules-module.html 上一篇讲了模块是如何被寻找到然后加载进来的,这篇则 ...

  3. 基于nodejs的终端天气查询

    国际惯例,先上效果图 前天,突然想到,怎么直接在命令行查询天气呢?好的,那就写一个吧.然后就开始找城市.天气的api接口,最终做出来这么一个东西. 安装方法:$ npm install tianqi ...

  4. dev gridcontrol纵向合并单元格设置

    1.要设置gridcontrol中指定列(columns中选中指定列)的AllowMerge属性为true; 2.要设置gridview中AllowCellMerge的属性为true; 3.如果只合并 ...

  5. background-position 50% 50%是如何计算的

    background-position:value1 value2 value1和value2的值可以值绝对值也可以是百分数,大部分值都很好理解,但是50% 50%这两个值是如何计算的呢? 图片水平和 ...

  6. C# Rotating Oval

    This program is used to show how to generate an oval. The moon's orbit around the sun is an oval two ...

  7. Android Bundle

    #Bundle类介绍 Bundle主要用于传递数据:它保存的数据,是以key-value(键值对)的形式存在的. 我们经常使用Bundle在Activity之间传递数据,传递的数据可以是boolean ...

  8. 我的一个小作品 android App ---校园资讯助手

        软件主界面采用Fragment+ViewPager组成.在点开后将会自动对学校新闻页面使用URl类来抓取,然后对网页中的信息提取,使用WebView来loadData在主界面上面显示, 为了使 ...

  9. 设计模式之UML类图的常见关系

    设计模式之UML类图的常见关系 本文来自转载 烧点饭博客 本篇会讲解在UML类图中,常见几种关系: 泛化(Generalization),依赖(Dependency),关联(Association), ...

  10. [转]扩展RBAC用户角色权限设计方案

    原文地址:http://www.iteye.com/topic/930648 RBAC(Role-Based Access Control,基于角色的访问控制),就是用户通过角色与权限进行关联.简单地 ...