题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4063 Description You are playing a flying game. In the game, player controls an aircraft in a 2D-space. The mission is to drive the craft from starting point to terminal point. The craft needs wireless s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4069 Problem Description Today we play a squiggly sudoku, The objective is to fill a 9*9 grid with digits so that each column, each row, and each of the nine Connecting-sub-grids that compose the grid co…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4064 Problem Description Carcassonne is a tile-based board game for two to five players.Square tiles are printed by city segments,road segments and field segments. The rule of the game is to put the tile…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4031 Problem Description Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack American again. However, American is protected by a high wall this time, which can be treating…

HDU 4291 A Short problem(2012 ACM/ICPC Asia Regional Chengdu Online) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=4291 Description 给一个式子求结果.类似Fibonacci的公式g(n)=3*g(n-1)+g[n-2]. Input 给你n(1<=n<=1e18) Output 求g(g(g(n))) Sample Input 样例第一个就是0什么鬼,虽然没影响.…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5492 题目大意: 一个N*M的矩阵,一个人从(1,1)走到(N,M),每次只能向下或向右走.求(N+M-1)ΣN+M-1(Ai-Aavg)2最小.Aavg为平均值. (N,M<=30,矩阵里的元素0<=C<=30) 题目思路: [动态规划] 首先化简式子,得原式=(N+M-1)ΣN+M-1(Ai2)-(ΣN+M-1Ai)2 f[i][j][k]表示走到A[i][j]格子上,此时前i+j-1…

Weak Pair Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 439    Accepted Submission(s): 155 Problem Description You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a…

与其说这是一次重温AC自动机+dp,倒不如说这是个坑,而且把队友给深坑了. 这个题目都没A得出来,我只觉得我以前的AC自动机的题目都白刷了——深坑啊. 题目的意思是给你两个串,每个串只含有R或者D,要求有多少种长为(n+m)的串(其中有n个R,m个D)同时含有这两个串作为子串.(我就不说题目描述了) 看完了题目我居然一开始用排列组合去做,搞了近一个小时,深坑队友啊. 其实正解是用AC自动机来dp出解的. 什么意思呢?把两个给定的串放到AC字典树上,并且弄好fail指针,变为tire图,构造好AC…

题目链接: Hdu 5459 Jesus Is Here 题目描述: s1 = 'c', s2 = 'ff', s3 = s1 + s2; 问sn里面所有的字符c的距离是多少? 解题思路: 直觉告诉我们,sn肯定由sn-1与sn-2推导出来的.然后呢,我们可以看出 n%2==1 的时候 sn-1 与 sn-2 由 ffff 衔接起来的,n%2==0 的时候,sn-1 与 sn-2由 ff 衔接起来的.告诉队友后,队友就把这个当成重要依据推啊,推啊!!到最后感觉丢队友自己看药丸,放弃02回来和队友…

思路:广搜, 因为空格加上动物最多只有32个那么对这32个进行编号,就能可以用一个数字来表示状态了,因为只有 ‘P’   'S' 'M' '.' 那么就可以用4进制刚好可以用64位表示. 接下去每次就是模拟了. 注意:  ‘S’ 不是只有一个. 一个东西如果不是'P'在动的话要先判断周围有没有‘P’,有的话要先吃掉      'P'在动的时候如果一个位置周围有多个东西,都要吃掉. #include<iostream> #include<cstdio> #include<alg…