hdu4745(区间dp,求最大回文串的长度)
来源:互联网 发布:数据预处理 去除噪声 编辑:程序博客网 时间:2024/06/10 15:55
Two Rabbits
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1346 Accepted Submission(s): 669
Problem Description
Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny afternoon, they planned to play a game with some stones. There were n stones on the ground and they were arranged as a clockwise ring. That is to say, the first stone was adjacent to the second stone and the n-th stone, and the second stone is adjacent to the first stone and the third stone, and so on. The weight of the i-th stone is ai.
The rabbits jumped from one stone to another. Tom always jumped clockwise, and Jerry always jumped anticlockwise.
At the beginning, the rabbits both choose a stone and stand on it. Then at each turn, Tom should choose a stone which have not been stepped by itself and then jumped to it, and Jerry should do the same thing as Tom, but the jumping direction is anti-clockwise.
For some unknown reason, at any time , the weight of the two stones on which the two rabbits stood should be equal. Besides, any rabbit couldn't jump over a stone which have been stepped by itself. In other words, if the Tom had stood on the second stone, it cannot jump from the first stone to the third stone or from the n-the stone to the 4-th stone.
Please note that during the whole process, it was OK for the two rabbits to stand on a same stone at the same time.
Now they want to find out the maximum turns they can play if they follow the optimal strategy.
The rabbits jumped from one stone to another. Tom always jumped clockwise, and Jerry always jumped anticlockwise.
At the beginning, the rabbits both choose a stone and stand on it. Then at each turn, Tom should choose a stone which have not been stepped by itself and then jumped to it, and Jerry should do the same thing as Tom, but the jumping direction is anti-clockwise.
For some unknown reason, at any time , the weight of the two stones on which the two rabbits stood should be equal. Besides, any rabbit couldn't jump over a stone which have been stepped by itself. In other words, if the Tom had stood on the second stone, it cannot jump from the first stone to the third stone or from the n-the stone to the 4-th stone.
Please note that during the whole process, it was OK for the two rabbits to stand on a same stone at the same time.
Now they want to find out the maximum turns they can play if they follow the optimal strategy.
Input
The input contains at most 20 test cases.
For each test cases, the first line contains a integer n denoting the number of stones.
The next line contains n integers separated by space, and the i-th integer ai denotes the weight of the i-th stone.(1 <= n <= 1000, 1 <= ai <= 1000)
The input ends with n = 0.
For each test cases, the first line contains a integer n denoting the number of stones.
The next line contains n integers separated by space, and the i-th integer ai denotes the weight of the i-th stone.(1 <= n <= 1000, 1 <= ai <= 1000)
The input ends with n = 0.
Output
For each test case, print a integer denoting the maximum turns.
Sample Input
1141 1 2 162 1 1 2 1 30
Sample Output
145HintFor the second case, the path of the Tom is 1, 2, 3, 4, and the path of Jerry is 1, 4, 3, 2.For the third case, the path of Tom is 1,2,3,4,5 and the path of Jerry is 4,3,2,1,5.
题意:两只兔子,在n块围成一个环形的石头上跳跃,每块石头有一个权值ai,一只从左往右跳,一只从右往左跳,每跳一次,两只兔子所在的石头的权值都要相等,在一圈内(各自不能超过各自的起点,也不能再次回到起点)问它们最多能经过多少个石头。
思路:其实仔细观察后可以发现其实就是求回文串的最大长度,只是最后处理的时候要注意的是因为是一个环,所以可能出现回文串+1的情况,所以得出答案的时候要进行一下判断。
#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>using namespace std;typedef long long ll;int dp[2010][2010];int x[2010];int fdp(int a,int b){ if(dp[a][b]!=-1) return dp[a][b]; if(a==b) return dp[a][b]=1; if(a>b) return dp[a][b]=0; int max0=0; max0=max(fdp(a+1,b),fdp(a,b-1)); if(x[a]==x[b]) max0=max(max0,fdp(a+1,b-1)+2); return dp[a][b]=max0;}int main(){ int n; while(~scanf("%d",&n)&&n) { memset(dp,-1,sizeof(dp)); for(int i=0; i<n; i++) scanf("%d",&x[i]),x[i+n]=x[i]; fdp(0,n*2-1); int ans=0; for(int i=0; i<n; i++) ans=max(ans,dp[i][i+n-1]), ans=max(ans,dp[i][i+n-2]+1);///判断是否为回文串+1的情况 printf("%d\n",ans); } return 0;}
0 0
- hdu4745(区间dp,求最大回文串的长度)
- hdu4745 Two Rabbits(区间dp,回文串思想)
- hdu4745(区间DP)
- hdu4745 最长回文子序列(区间DP)
- 求最大回文子串的长度
- HDU4745环的回文串
- hdu4745——Two Rabbits(区间dp,最长回文子序列)
- hdu4745 Two Rabbits--最长不连续回文序列 & 区间dp
- HDU4745:Two Rabbits(区间DP)
- HDU4745 Two Rabbits(区间dp)
- hdu4745(区间dp)
- hdu4745区间dp
- 求字符串中最大的回文长度
- hdu4632(区间dp求字符串的回文子串数)
- UVa 11404 - Palindromic Subsequence(最大回文串,区间DP)
- Palindrome(Manacher求最大回文长度)
- poj 2955 Brackets(区间DP,经典问题)求有规律的括号的最大长度
- manacher求最大回文长度
- 猿题库 iOS 客户端架构设计
- 排序算法java 一 --快速排序、直接插入、希尔排序
- 【USACO题库】1.4.3 Arithmetic Progressions等差数列(模拟)
- 简单动态分配内存1
- PHP+JS动态显示当前时间
- hdu4745(区间dp,求最大回文串的长度)
- 简单动态分配内存2
- mysql之TIMESTAMP(时间戳)用法详解
- Spring中AutowireCapableBeanFactory的使用
- close和dispose的区别
- java JDK 配置后不生效竟然
- C++中 #ifdef 和#endif的作用
- 简单动态分配内存3
- Linux与Windows的不同