http://uoj.ac/problem/29 cdq四次处理出一直向左, 一直向右, 向左后回到起点, 向右后回到起点的dp数组,最后统计答案. 举例:\(fi\)表示一直向右走i天能参观的最多景点数. 其中有一个很重要的条件\(fi≤fi+1fi≤fi+1\),这个条件是分治的前提. 关于这个条件的证明,我想了好久才想出来,用反证法证明一下就行. 分治时需要用主席树维护路径上的前k大和. #include<cstdio> #include<cstring> #include&…

题目传送门 /* 水题:已知1928年1月1日是星期日,若是闰年加1,总天数对7取余判断就好了: */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set> #include <…

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5500 The 12th Zhejiang Provincial Collegiate Programming Contest - H May Day Holiday Time Limit: 2 Seconds      Memory Limit: 65536 KB As a university advocating self-learning and work…

May Day Holiday Time Limit: 2 Seconds Memory Limit: 65536 KB As a university advocating self-learning and work-rest balance, Marjar University has so many days of rest, including holidays and weekends. Each weekend, which consists of Saturday and Sun…

Holiday's Accommodation Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 200000/200000 K (Java/Others)Total Submission(s): 2009    Accepted Submission(s): 558 Problem Description Nowadays, people have many ways to save money on accommodation w…

Summer Holiday Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2256    Accepted Submission(s): 1050 Problem Description To see a World in a Grain of Sand And a Heaven in a Wild Flower, Hold Inf…

题目链接:  HDU 4118 Holiday's Accommodation 分析: 可以知道每条边要走的次数刚好的是这条边两端的点数的最小值的两倍. 代码: #include<iostream> #include<cstdio> #include<cstring> #include<stack> using namespace std; const int maxn=100000+10; struct node{ int to, dix, next; }…

Summer Holiday Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2319    Accepted Submission(s): 1082 Problem Description To see a World in a Grain of Sand And a Heaven in a Wild Flower, Hold Inf…

题目 题目链接 大意:从左到右有\(n\)个城市,一开始在城市\(start\),每一天有两种选择: 前往相邻的城市. 访问当前城市(每个城市只能访问一次),访问城市\(i\)可以获得\(attraction_i\)的分数. 问:在\(d\)天内最多能获得多少分数. 算法 首先,分成左右两边来做,路线有\(4\)种: 往右走 往左左 现往右再往左 先往左再往右 将左右分开做,对于往右走,计算出\(f(i)\),表示往右走不回头在\(i\)内最大得分:以及\(g(i)\),表示往右走但是要回到原点…

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3876 Description As a university advocating self-learning and work-rest balance, Marjar University has so many days of rest, including holidays and…