Josephina and RPG

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2013长沙赛区现场赛 J - Josephina and RPG

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…

hdu4800 Josephina and RPG 解题报告

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…

Josephina and RPG HDU - 4800

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…

DP ZOJ 3735 Josephina and RPG

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]][…

hdu 4800 Josephina and RPG

简单dp #include<cstdio> #define maxn 10005 #include<cstring> #include<algorithm> using namespace std; ][maxn]; ][]; int num[maxn]; int main() { int n,m; while(scanf("%d",&m)!=EOF) { m=(m*(m-)*(m-)/); ; i<m; i++) { ; j<m…

HDU 4800/zoj 3735 Josephina and RPG 2013 长沙现场赛J题

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…