A - Jessica's Reading Problem POJ - 3320 Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a…

Problem statement Given n items with size Ai and value Vi, and a backpack with size m. What's the maximum value can you put into the backpack? Solution 0/1 knapsack problem is a classical dynamic programming model. There is a knapsack with the capaci…

Jessica's Reading Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17562   Accepted: 6099 Description Jessica's a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent litt…

You must have heard of the Knight's Tour problem. In that problem, a knight is placed on an empty chess board and you are to determine whether it can visit each square on the board exactly once. Let's consider a variation of the knight's tour problem…

Jessica's Reading Problem 题目大意:Jessica期末考试临时抱佛脚想读一本书把知识点掌握,但是知识点很多,而且很多都是重复的,她想读最少的连续的页数把知识点全部掌握(知识点都在书上,每一页都是一个知识点) 这一题可以用3061的游标卡尺法,我们可以先数数书上倒到底有多少个知识点,因为知识点都是序数,我们可以用二分法直接找到O(PlogP),这里可以采用set模板直接偷懒了,然后我们就可以用游标卡尺法了,因为所有知识点都要出现一次,所以我们统计新的知识点的出现就好了,最…

Xuanxuan has n sticks of different length. One day, she puts all her sticks in a line, represented by S1, S2, S3, ...Sn. After measuring the length of each stick Sk (1 <= k <= n), she finds that for some sticks Si and Sj (1<= i < j <= n), e…

题目大意:有两块木板交叉起来接雨水,问最多能接多少.   分析:题目描述很简单,不过有些细节还是需要注意到,如下图几种情况:   #include<stdio.h> #include<math.h> #include<string.h> #include<algorithm> using namespace std; ; ; ; int sign(double val) { ; ; ; } struct Point { double x, y; Point(…

题意:n页书,然后n个数表示各个知识点ai,然后,输出最小覆盖的页数. #include<iostream> #include<cstdio> #include<set> #include<map> using namespace std; ; int num[maxn]; int main(){ int n; set<int>ss; map<int, int>mm; scanf("%d", &n); ;…

最大01子矩阵和,就是一个矩阵的元素不是0就是1,然后求最大的子矩阵,子矩阵里的元素都是相同的. 这个题目,三个oj有不同的要求,hoj的要求是5s,poj是3秒,hdu是1秒.不同的要求就对应不同的难度,不同的逼格. 先看最low的, HOJ 1664 5秒钟的时间,够长了.我很容易想到可以最大子矩阵和来求解,二者本来就很像,关于最大子矩阵和这个博客里有介绍 最大子矩阵和 这里我们可以把F变成1,把R变成负无穷大,这样求解最大子矩阵和就可以得到答案 #include <iostream> #…

Stars Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 42898   Accepted: 18664 Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a st…