【HDU】4405 Aeroplane chess】的更多相关文章

http://acm.hdu.edu.cn/showproblem.php?pid=4405 题意:每次可以走1~6格,初始化在第0格,走到>=n的格子就结束.还有m个传送门,表示可以从X[i]格传送到Y[i]而不需要消耗次数,X[i]<Y[i].n<=100000, m<=1000. #include <cstdio> #include <cstring> using namespace std; double d[100010]; int n, m, m…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4405 Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) 问题描述 Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz s…
传送门:pid=4888">[HDU]4888 Redraw Beautiful Drawings 题目分析: 比赛的时候看出是个网络流,可是没有敲出来.各种反面样例推倒自己(究其原因是不愿意写暴力推断的).. 首先是简单的行列建边.源点向行建边.容量为该行元素和,汇点和列建边.容量为该列元素和.全部的行向全部的列建边,容量为K. 跑一次最大流.满流则有解,否则无解. 接下来是推断解是否唯一. 这个题解压根没看懂.还是暴力大法好. 最简单的思想就是枚举在一个矩形的四个端点.设A.D为主对角…
Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces a…
Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1667    Accepted Submission(s): 1123 Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids lab…
Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1503    Accepted Submission(s): 1025 Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids la…
Problem Description Hzz loves aeroplane chess very much. The chess map contains N+ grids labeled to N. Hzz starts at grid . For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are ,,,,,)…
Aeroplane chess Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the number…
原题目:悼念512汶川大地震遇难同胞——珍惜现在,感恩生活 [算法]多重背包(有限背包) 动态规划 [题解]http://blog.csdn.net/acdreamers/article/details/8563283 优化:若物品数量(num[i])*物品重量(w[i])>背包容量(m),就相当于无限背包. 对于num[i],可以拆成若干个01背包来实现1...num[i]的全覆盖,二进制原理: 1...k中的数可以由1.2.4...2t.k-2t+1+1(2t+2 > k ≥ 2t+1)组…
[题目]2017"百度之星"程序设计大赛 - 初赛(A) [题意]给定n个点的带边权树,m条编号1~m的路径,Q次询问编号区间[L,R]所有链的交集的长度.n<=500000. [算法]线段树+RMQ-LCA+树链的交 [题解]树链的交:记一条链为(a1,b1),LCA为c1.另一条链为(a2,b2),LCA为c2.记a1a2,a1b2,b1a2,b1b2的LCA为d1,d2,d3,d4,按深度排序后得deep[d1]<=deep[d2]<=deep[d3]<=…