题目传送门 题意:告诉你C(m,3)个队伍相互之间的胜率,然后要你依次对战n个AI队伍,首先任选一种队伍,然后战胜一个AI后可以选择替换成AI的队伍,也可以不换,问你最后最大的胜率是多少. 分析:dp[i][j][0/1] 表示第i个AI,用j的id去攻打,此j可以是上一个状态交换AI的id而来也可以不是,状态转移方程: dp[i][j][0] = max (dp[i-1][j][0], dp[i-1][j][1]) * p[j][a[i]]; if (i > 1) dp[i][a[i-1]][…

第一年参加现场赛,比赛的时候就A了这一道,基本全场都A的签到题竟然A不出来,结果题目重现的时候1A,好受打击 ORZ..... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4800 题目大意:给定C(3,N)支队伍之间对战的获胜概率,再给定一个序列存放队伍编号,每次获胜之后可以选择和当前战胜的对手换队伍.问按给定序列依次挑战全部胜利的最大概率. 解题思路:状压DP dp[i][j]表示使用队伍i从编号j开始挑战全胜的概率,ai[i]表示i位置的队…

J - Josephina and RPG Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4800 Description A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles of cha…

Josephina and RPG Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 929    Accepted Submission(s): 265Special Judge Problem Description A role-playing game (RPG and sometimes roleplaying game) is…

Josephina and RPG Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 511    Accepted Submission(s): 139Special Judge Problem Description A role-playing game (RPG and sometimes roleplaying game) is…

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3735 好久没做DP题了,一开始没理解题目里的C(M,3)是干什么,原来就是组合,C M 取3,就等于n*(n-1)*(n-2)/6;题目里还有一个细节是说电脑玩家是要一个接着一个打败,这样,规划方向也确定了,设dp[i][j]为当前打败了Ai电脑,并且阵容为j的概率最大值,dp[i][j]=max(dp[i][j],dp[i-1][j]*p[j][no[i]]) ,p[]…

A role-playing game (RPG and sometimes roleplaying game) is a game in which players assume the roles of characters in a fictional setting. Players take responsibility for acting out these roles within a narrative, either through literal acting or thr…

题目传送门 /* DP:dp 表示当前输入的x前的包含x的子序列的和, 求和方法是找到之前出现x的位置(a[x])的区间内的子序列: sum 表示当前输入x前的所有和: a[x] 表示id: 详细解释:http://blog.csdn.net/u013050857/article/details/45285515 */ #include <cstdio> #include <algorithm> #include <cmath> #include <iostrea…

Help Me Escape Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submit Status Practice ZOJ 3640 Appoint description:  System Crawler  (2014-10-22) Description Background     If thou doest well, shalt thou not be accepted? an…

题目传送门 题意:要求任意连续子序列中0和1的数量差不超过k的方案数 分析:想好状态其实不难.dp[i][j][k]表示考虑前i长度,后缀中最大的 sum(0) - sum(1) = j, sum (1) - sum (0) = k的方案数,合并以下可以得到最大的|sum(0) - sum(1)| = j + k,所以j+k <= K,最后考虑当前i放0或1就可以转移状态了. #include <bits/stdc++.h> using namespace std; typedef lo…