HDU 4745 Two Rabbits 区间dp_回文序列

题目链接：

http://blog.csdn.net/scnu_jiechao/article/details/11759333

Two Rabbits

Time Limit: 10000/5000 MS (Java/Others)
Memory Limit: 65535/65535 K (Java/Others)
#### 问题描述
> 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.
#### 输入
> 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 The input ends with n = 0.
#### 输出
> For each test case, print a integer denoting the maximum turns.
####样例输入
> 1
> 1
> 4
> 1 1 2 1
> 6
> 2 1 1 2 1 3
> 0

样例输出

1

4

5

题意

给你一个环，一个人任选一点顺时针跳，另一个人也任选一点逆时针跳，两个人的起点可以相同，每回合每个人向前条一次，现在跳的时候有两个要求：1、两个人每次跳的位置上的数字必须相同，2、跳的最远位置不能超过一圈。

求最大的回合数

题解

Portal

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef int LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=2222;

int n;
int a[maxn];
int dp[maxn][maxn];

int main() {
    while(scf("%d",&n)==1&&n){
        for(int i=0;i<n;i++) scf("%d",&a[i]);
//        rep(i,0,len) prf("%d ",a[i]); puts("");

        clr(dp,0);
        for(int r=0;r<n;r++){
            for(int l=r;l>=0;l--){
                if(l==r) dp[l][r]=1;
                else{
                    if(a[l]==a[r]){
                        dp[l][r]=max(dp[l][r],dp[l+1][r-1]+2);
                    }
                    dp[l][r]=max(dp[l][r],dp[l][r-1]);
                    dp[l][r]=max(dp[l][r],dp[l+1][r]);
                }
            }
        }
        int ans=dp[0][n-1];
        for(int i=0;i<n-1;i++) ans=max(ans,dp[0][i]+dp[i+1][n-1]);

        prf("%d\n",ans);
    }
    return 0;
}

//end-----------------------------------------------------------------------