传说的SB DP: 题目 Problem Description Lee has a string of n pearls. In the beginning, all the pearls have no color. He plans to color the pearls to make it more fascinating. He drew his ideal pattern of the string on a paper and asks for your help. In eac…

Mart Master II Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 675    Accepted Submission(s): 237 Problem Description Trader Dogy lives in city S, which consists of n districts. There are n - 1…

03 hdu5009 状态转移方程很好想,dp[i] = min(dp[j]+o[j~i]^2,dp[i]) ,o[j~i]表示从j到i颜色的种数. 普通的O(n*n)是会超时的,可以想到o[]最大为sqrt(n),问题是怎么快速找到从i开始往前2种颜色.三种.四种...o[]种的位置. 离散化之后,可以边走边记录某个数最后一个出现的位置,初始为-1,而所要求的位置就等于 if(last[a[i]]==-1) 该数没有出现过,num[i][1] = i,num[i][j+1] = num[i-1…

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…

题目链接 A题:(字符串查找,水题) 题意 :输入字符串,如果字符串中包含“ Apple”, “iPhone”, “iPod”, “iPad” 就输出 “MAI MAI MAI!”,如果出现 “Sony” 就输出“SONY DAFA IS GOOD!” ,大小写敏感. 思路 : 字符串查找,水题. #include <string.h> #include <stdio.h> #include <iostream> using namespace std ; ]; int…

Clone Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other) Total Submission(s) : 8   Accepted Submission(s) : 5 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description After eating food from Chernobyl,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5029 Problem Description The soil is cracking up because of the drought and the rabbit kingdom is facing a serious famine. The RRC(Rabbit Red Cross) organizes the distribution of relief grain in the disa…

Problem Description After eating food from Chernobyl, DRD got a super power: he could clone himself right now! He used this power for several times. He found out that this power was not as perfect as he wanted. For example, some of the cloned objects…

Tree http://acm.hdu.edu.cn/showproblem.php?pid=5044 树链剖分,区间更新的时候要用on的左++右--的标记方法,要手动扩栈,用c++交,综合以上的条件可过. #include<cstdio> #include<cstring> #include<algorithm> #pragma comment(linker, "/STACK:36777216") #define mt(a,b) memset(a,…

Wang Xifeng's Little Plot http://acm.hdu.edu.cn/showproblem.php?pid=5024 预处理出每个点八个方向能走的最远距离,然后枚举起点,枚举方向,每走一步都要枚举左转和右转的情况,因为预处理好了,所以可以直接算出来. #include<cstdio> #include<algorithm> using namespace std; ; char a[M][M]; ]; ,-,,,,,,-}; ,,,,,-,-,-}; b…