转自:http://blog.csdn.net/accelerator_/article/details/39271751 吐血ac... 11668627 2014-09-16 22:15:24 Accepted 5009 1265MS 1980K 2290 B G++ czy   Paint Pearls Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Subm…
Paint Pearls 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 e…
Paint Pearls 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 ea…
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 each operation,…
首先把具有相同颜色的点缩成一个点,即数据离散化. 然后使用dp[i]表示涂满前i个点的最小代价.对于第i+1个点,有两种情况: 1)自己单独涂,即dp[i+1] = dp[i] + 1 2)从第k个节点之后(不包括k)到第i+1个节点一次涂完,且一起涂的节点共有num种颜色,即dp[i+1] = dp[k] + num * num 从而可以得到状态转移方程dp[i+1] = min(dp[i], dp[k] + num * num) 但是如果从后往前遍历每一个k,会超时. 因此我们可以使用双向链…
Ellipsoid Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 850    Accepted Submission(s): 271 Special Judge Problem Description Given a 3-dimension ellipsoid(椭球面) your task is to find the minima…
太吊了,反正我不会 /* HDU 4035 dp求期望的题. 题意: 有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树, 从结点1出发,开始走,在每个结点i都有3种可能: 1.被杀死,回到结点1处(概率为ki) 2.找到出口,走出迷宫 (概率为ei) 3.和该点相连有m条边,随机走一条 求:走出迷宫所要走的边数的期望值. 设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望.E[1]即为所求. 叶子结点: E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[fa…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=5014 题意:给出范围N,给出0-N的一个排列a.让你求出另外一个排列b,使 t = a1 ^ b1 + a2 ^ b2 + ...+an ^ bn(^表示异或)最大.并求出最大的t. 思路:首先,我们要注意到的是,一定存在解,且解不唯一.同时,因为是异或操作,要想得到最大值,我们必须让ai,bi对应的二进制表示,0和1错开,这样才不会使值减少. 这样,我们要构造出错开的0和1的二进制表达即可. #inc…
专题训练 #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; ; const int INF=0x3fffffff; struct Node { int to; int next; }edge[MAXN*]; int tol; int head[MAXN]; ]; ]; ]; void init() { memset(…
题目:给出1-n连续的方格,从0开始,每一个格子有4个状态,左右脚交替,向右跳,而且每一步的步长必须在给定的区间之内.当跳出n个格子或者没有格子可以跳的时候就结束了,求出游戏的期望步数 0:表示不能到达这个格子 1:表示左脚跳进这个格子 2:表示右脚跳进这个格子 3:随意哪个脚跳进这个格子,而且下一步随意用哪个脚 dp[i][j] :表示走到第 i 个格子在 j 状态的期望. 当j=1时,你可以走到dp[i+k][2],dp[i+k][3], 当j=2时,你可以走到dp[i+k][1],dp[i…