Brute Force Sorting Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1043    Accepted Submission(s): 272 Problem Description Beerus needs to sort an array of N integers. Algorithms are not Beerus…

http://acm.hdu.edu.cn/showproblem.php?pid=6215 题意:给出一个序列,对于每个数,它必须大于等于它前一个数,小于等于后一个数,如果不满足,就删去.然后继续去判断剩下的数,直到最后都满足. 思路: 建立双向链表,如果一个数是需要删除的,那么它只会影响它前一个的数和后一个的数,这样只需要把它前面的数保存下来,下次再跑即可. #include<iostream> #include<algorithm> #include<cstring&g…

题目链接 Problem Description Beerus needs to sort an array of N integers. Algorithms are not Beerus's strength. Destruction is what he excels. He can destroy all unsorted numbers in the array simultaneously. A number A[i] of the array is sorted if it sat…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6215 题解:类似双链表的模拟. #include <iostream> #include <cstring> #include <cstdio> using namespace std; ; int a[M] , Next[M] , pre[M] , que[M]; int main() { int t; scanf("%d" , &t); wh…

一层一层删 链表模拟 最开始写的是一个一个删的 WA #include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define pb push_back //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<…

Brute Force Sorting Time Limit: 1 Sec  Memory Limit: 128 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=6215 Description Beerus needs to sort an array of N integers. Algorithms are not Beerus's strength. Destruction is what he excels. He can destr…

A simple brute force problem. Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on HDU. Original ID: 497164-bit integer IO format: %I64d      Java class name: Main There's a company with several projects to be done. Finish a projec…

题目来源:http://community.topcoder.com/stat?c=problem_statement&pm=12581 Burte Force 算法,求解了所有了情况,注意 next_permutation 函数的用法. #include <iostream> #include <vector> #include <limits> #include <algorithm> using namespace std; class Col…

题目来源:http://community.topcoder.com/stat?c=problem_statement&pm=3005&rd=5858 思路: 如果直接用Brute Force搜索所有可能的圆的话,那么搜索空间将很大,所以我用了一个priority_queue结构,将搜索的顺序变为按圆的半径从大到小搜索,所以当搜索到符合条件的圆时,即可停止搜索.这样可以大大减少搜索范围,不过对于最坏的情况,也就是没有符合条件的圆时,还是会将所有的可能情况都搜索到. 代码如下: #inclu…

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6215 题目: Brute Force Sorting Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 496    Accepted Submission(s): 119 Problem Description Beerus need…