HDU 4745 Two Rabbits

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）它们最多能经过多少个石头(1 <= n <= 1000, 1 <= ai <= 1000)。

分析：
其实就是求一个环中，非连续最长回文子序列的长度。

dp[i][j] = max{ dp[i + 1][j], d[i][j - 1], (if a[i] == a[j]) dp[i + 1][j - 1] + 2 }

直接当成一个链求dp，然后把链切成两半，求出两边的回文长度，最大的和就是解。这里不用考虑起点问题，因为两边的回文中点都可以做起点。

Two Rabbits

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 941    Accepted Submission(s): 495

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.

 

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.

 

Output

For each test case, print a integer denoting the maximum turns.

 

Sample Input

   
   
   
   

    
    
    
    1
1
4
1 1 2 1
6
2 1 1 2 1 3
0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
4
5

    
    
    
    
 
     
 
      Hint 
     
 For 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. 
    
    
    
    
     
   
   
   
   

 

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>

using namespace std;
#define ll long long
#define prt(k) cerr<<#k" = "<<k<<endl

const int N = 2014;
int dp[N][N];
int a[N];
int main()
{
    int n;
    while(cin>>n,n)
    {
        for(int i=1;i<=n;i++) scanf("%d", a+i);
        memset(dp,0,sizeof dp);
        for(int i=1;i<=n;i++) dp[i][i] = 1;
        for(int len=2;len<=n;len++)
        {
            for(int i=1;i+len-1<=n;i++)
            {
                int j=i+len-1;
                dp[i][j]=max(dp[i][j-1], dp[i+1][j]);
                if(a[i]==a[j]) dp[i][j]=max(dp[i][j], dp[i+1][j-1] + 2);
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++)ans=max(ans,dp[1][i]+dp[i+1][n]);
        printf("%d\n",ans);
    }
}

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
#define prt(k) cout<<#k"="<<k<<endl;
const int inf = 0x3f3f3f3f;
struct P
{
	int x,y;
	P() {}
	P(int xx, int yy) { x=xx, y=yy; }
}p[33];
int n;
struct Mat
{
	P p[4];
	int maxx, minx, maxy, miny;
	int S;
	Mat() {}
	Mat(P a, P b, P c, P d)
	{
        p[0] = a; p[1] = b; p[2] = c; p[3] = d;
        maxx = maxy = -inf;
		minx = miny = inf;
		for(int i=0;i<4;i++)
        {
            maxx = max(maxx, p[i].x);
            maxy = max(maxy, p[i].y);
            minx = min(minx, p[i].x);
            miny = min(miny, p[i].y);
        }
    	S = (maxy - miny) * (maxx - minx);
    }
	bool in(P a)
	{
	    return minx<=a.x&&a.x<=maxx && miny<=a.y&&a.y<=maxy;
	}
	bool inside(P a)
	{
	    return minx<a.x&&a.x<maxx && miny<a.y&&a.y<maxy;
	}
};

vector<Mat> mat;
int jiao(Mat a, Mat b)
{
    int c1=0; int c2=0;
    int t1 = 0;
    int t2 = 0;
    for(int i=0;i<4;i++) {
        if(b.inside(a.p[i])) c1++;
        if(a.inside(b.p[i])) c2++;
        if(b.in(a.p[i])) t1++;
        if(a.in(b.p[i])) t2++;
    }
    int t ;
    if((c1==4) || (c2==4)) t = max(a.S, b.S);
    else {
        if(t1==0 && t2==0) { t = a.S + b.S;  }
        else t =  -1;
    }
    return t;
}

int main()
{
	while(cin>>n, n)
	{
		for(int i=0;i<n;i++) cin>>p[i].x>>p[i].y;
		mat.clear();
		for(int a=0;a<n;a++)
			for(int b=a+1;b<n;b++)
				if(p[a].x == p[b].x)
				for(int c=0;c<n;c++)
					if(p[a].y == p[c].y && a != c && b != c)
						for(int d=0; d<n; d++)
							if(p[d].y == p[b].y && p[d].x == p[c].x)
							{
								Mat t = Mat(p[a],p[b],p[c],p[d]);
								if(t.S==0) continue;
								mat.push_back(t);
							}
		int m = mat.size();
		int ans = -1;
		for(int i=0;i<m;i++)
            for(int j=i+1;j<m;j++)
        {
            int tmp = jiao(mat[i], mat[j]);
            ans = max(ans, tmp);
        }
        if(ans == -1) puts("imp");
        else printf("%d\n", ans);
	}
	return 0;
}