POJ 1887Testingthe CATCHER (LIS:最长下降子序列) http://poj.org/problem?id=3903 题意: 给你一个长度为n (n<=200000) 的数字序列, 要你求该序列中的最长(严格)下降子序列的长度. 分析:        读取全部输入, 将原始数组逆向, 然后求最长严格上升子序列就可以. 因为n的规模达到20W, 所以仅仅能用O(nlogn)的算法求.        令g[i]==x表示当前遍历到的长度为i的全部最长上升子序列中的最小序列末…

Testing the CATCHER Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 16515 Accepted: 6082 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile called th…

Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13396   Accepted: 4905 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile calle…

/* 简单dp,要记录顺序 解:先排序,然后是一个最长下降子序列 ,中间需记录顺序 dp[i]=Max(dp[i],dp[j]+1); */ #include<stdio.h> #include<string.h> #include<stdlib.h> #define N 1100 /*w,s代表重量和速度,index记录原来输入时的顺序下标,pre指向排序后的上一个下标,answer记录排序后每一个位置的最优值*/ typedef struct node { int…

https://ac.nowcoder.com/acm/contest/3007/C 将木板按照Xi从小到大排序,将这时的Yi数列记为Zi数列,则问题变成将Zi划分为尽可能少的若干组上升子序列. 根据Dilworth定理,最小组数等于Zi的最长下降子序列长度. 要求最长下降子序列的长度,我们有一种经典的二分优化dp的方法,在这里不再详述. 借助这种做法我们能给出一种构造方法,在求出最小组数的同时得出方案. 将状态数组的每个位置变为栈,用入栈操作代替修改元素操作,即可在求出组数的同时,用这些栈来完…

一.Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile called the CATCHER which is capable of intercepting multiple incoming offensive missiles. The CATCHER is su…

Language: Default Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15207   Accepted: 5595 Description A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defen…

Bridging signals Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9234   Accepted: 5037 Description 'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up co…

题目链接:http://poj.org/problem?id=3903 题目链接:http://poj.org/problem?id=1631 题目链接:http://poj.org/problem?id=1887 题目解析: 这两道题都是直接求最长上升子序列,没什么好说的. POJ 3903这题n为1000000,如果用n^2的算法肯定超时,所以要选择nlogn的算法.都是简单题. #include <iostream> #include <string.h> #include…

题意:题目太长没看,直接看输入输出猜出是最长下降子序列 用了以前的代码直接a了,做法类似贪心,把最小的顺序数存在数组里面,每次二分更新数组得出最长上升子序列 #include<iostream> #include<cstdio> using namespace std; int main() { int dp[40002],a[40002],n,t,i,low,up,top,mid,max,tmp,k,b[40002],cas=1; while(1) { scanf("%…