Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 10575   Accepted: 5489 Description BackgroundRaymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just b…

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…

Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7787   Accepted: 4247 Description Background Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just b…

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,分别求子序列的逆序数:对于右子序列中的每一个数,求出左序列中大于它的数的数目,计算的和即为解. 另外,使用Merge排序时,可以很容易求得对于右子序列中的每一个数,左序列中大于它的数的数目. 代码如下: #include <stdio.h> #include <limits.h> ; + ; long…

点击打开链接 Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7810   Accepted: 4261 Description Background  Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instan…

Brainman Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 12175   Accepted: 6147 Description Background Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just…

[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, 通过子问题的答案反推得到总的答案 通过归并排序求逆序对数的算法流程图如下 : [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <…

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,我们知道,求逆序对最典型的方法就是树状数组,但是还有一种方法就是Merge_sort(),即归并排序.实际上归并排序的交换次数就是这个数组的逆序对个数,归并排序是将数列a[l,h]分成两半a[l,mid]和a[mid+1,h]分别进行归并排序,然后再将这两半合并起来.在合并的过程中(设l<=i<=mid…

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, 但是有编码难度.. 二叉排序树,内存巨大,时间复杂度高,但是非常好写.. 归并排序版本: #include <stdio.h> #include <string.h> long long merge_arr(int arr[], int first, int mid, int last, i…

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? 解题思路: 根据冒泡排序的特点,我们可知,本题只需要统计每一个数的逆序数(如果有i<j,存在a[i] > a[j],则称a[i]与 a[j]为逆序数对),输出所有的数的逆序数的和用普通排序一定会超时,但是比较快的排序,像快排又无法统计 交换次数,这里就很好地体现了归并排序的优点.典型的利用归并排序求逆…