Largest Rectangle in a Histogram Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15396    Accepted Submission(s): 4470 Problem Description A histogram is a polygon composed of a sequence of rect…

E - Largest Rectangle in a Histogram Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1506 Appoint description: Description A histogram is a polygon composed of a sequence of rectangles aligned a…

一看就是状压,由于是类似博弈的游戏.游戏里的两人都是绝对聪明,那么先手的选择是能够确定最终局面的. 实际上是枚举最终局面情况,0代表是被Bob拿走的,1为Alice拿走的,当时Alice拿走且满足变换成魔法石,那么相当于是Alice完成了该次操作,增加上次状态值,否则相当于先后手交换,该状态下减去上个状态值. /** @Date : 2017-09-12 23:51:53 * @FileName: HDU 4778 状压DP 或 记忆化 好题.cpp * @Platform: Windows *…

HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 1005 using namespace std; int v[nmax],w[nmax],dp[nmax]; int main() { //freopen("in…

Bomb HDU - 3555 (数位DP) The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49&…

/* 多谢了“闭眼,睁眼” 同学给我一套dp专题,不然真是没接触过这种题型. 做个4个简单的,很高兴有所收获. 2013-08-06 /* HDU 1506 最基础的一道题目,其主要精髓就在于两个数组 l[i],r[i]; 其中,l[i]用来存储第i个矩形的左边界,r[i]存储的是第i个矩形的右边界,也就是说对于任意的 l[i]<=x<=r[i]都有a[x]>=a[i]  ; #include<stdio.h> #include<string.h> #define…

Largest Rectangle in a Histogram Problem Description: A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left sh…

题目链接:  HDU 1011 树形背包(DP) Starship Troopers 题意:  地图中有一些房间, 每个房间有一定的bugs和得到brains的可能性值, 一个人带领m支军队从入口(房间1)进入, 只有到达某个房间并且将bugs全部杀死, 才能得到相应该的值. 问最多能获得多少可能性值. PS  1). 一支军队能杀死 20 bugs,  当一支军队发生战争之后就不能再到其它地方去了 2) . 不能走回头路 分析:  [树形背包] 用dp[i][j]表示到达房间 i 的军队数为…

职务地址:HDU 1950 这题是求最长上升序列,可是普通的最长上升序列求法时间复杂度是O(n*n).显然会超时.于是便学了一种O(n*logn)的方法.也非常好理解. 感觉还用到了一点贪心的思想. 详细的见这篇博客吧,写的非常通俗易懂.传送门 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h>…

Ring Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3180    Accepted Submission(s): 1033 Problem Description For the hope of a forever love, Steven is planning to send a ring to Jane with a rom…