题目传送门
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
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)。
题解：
- 其实就是求一个环中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 }
但是，这个dp公式仅仅是求出一个序列的非连续最长回文子序列，题目的序列是环状的，有两种思路 - 解法一： 将环倍增成链，求出窗口为n的最长子序列，但这不是最终的解，你可以试看看Sample 2,是只能得出4,因为它在选中的回文外面还可以选中一个当做起点来跳，所以外面得判断找出来的回文外面是否还有可以当起点的石头，即可以找窗口为(n-1)的长度+1。所以解即找 窗口为n的长度或者 窗口为(n-1)的长度+1 的最大值。
- 解法二： 我们看看样例：2 1 1 2 1 3 答案是 5，即 2 1 1 2 1 3，其实就是 [1,4] 区间的最长回文加上 [5,6] 区间的最长回文，因此我们枚举中间点 i ，求出 max(dp[1,i] + dp[i+1,n]) 就是最终结果
AC-Code
- 解法一
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <fstream>
#include <utility>
using namespace std;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;const int INF = 0x7fffffff;
const int MAXN = 1001 << 1;
const int MOD = 1e9 + 7;int a[MAXN];
int dp[MAXN][MAXN];int main() {
int n;while (cin >> n, n) {
memset(dp, 0, sizeof(dp));for (int i = 1; i <= n; i++) {
cin >> a[i];a[n + i] = a[i];dp[i][i] = 1;}for (int len = 1; len <= 2 * n; len++) {
for (int i = 1; i <= 2 * n; i++) {
int j = i + len;if (j > 2 * n)break;dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);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[i][n + i - 1]);for (int i = 1; i <= n; i++)ans = max(ans, dp[i][n + i - 2] + 1);cout << ans << endl;}return 0;
}
- 解法二
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <fstream>
#include <utility>
using namespace std;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long ll;const int INF = 0x7fffffff;
const int MAXN = 1001 << 1;
const int MOD = 1e9 + 7;int a[MAXN];
int dp[MAXN][MAXN];int main() {
int n;while (cin >> n, n) {
memset(dp, 0, sizeof(dp));for (int i = 1; i <= n; i++) {
cin >> a[i];dp[i][i] = 1;}for (int len = 1; len <= n; len++) {
for (int i = 1; i <= n; i++) {
int j = i + len;if (j > n)break;dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);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]);cout << ans << endl;}return 0;
}