一 原题
A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.
A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.
Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.
A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 ≤ ap2 ≤ ... ≤ apk. The length of the subsequence is k.
The first line contains an integer n (1 ≤ n ≤ 2000), denoting the length of the original sequence.
The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 ≤ ai ≤ 2, i = 1, 2, ..., n).
Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.
4 1 2 1 2
4
10 1 1 2 2 2 1 1 2 2 1
9
In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.
In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.
二 分析
题意:给定一个长度不超过2000的由1和2组成的字符串S,可以选定一个区间[l, r],将S的子串Si...Sj做一次reverse操作(变成Sj..Si),问怎么选定区间,使得的到的新串的最长非严格递增序列的长度最长。并输出其长度。
这题一开始我把reverse理解错了。。以为是把[l, r]中的1变成2,2变成1。但题意就是做std::reverse()操作。不难想到就是要求S中最长的满足1*2*1*2*的子序列。用动态规划解决,要求的子序列有4段:1*,1*2*,1*2*1*,1*2*1*2*,dp[i][j](i=1, 2, 3, 4. j = 1, 2, ..., S.length)表示S的前缀S1...Sj满足前i段要求的最长子序列长度。
三 代码
/*
PROB: cf 933A
LANG: c++
AUTHOR: maxkibble
*/
#include <cstdio>
#include <algorithm>
const int maxn = 2e3 + 10;
const int s[5] = {0, 1, 2, 1, 2};
int n, a[maxn], dp[5][maxn];
/* Description: find the longest 1*2*1*2* subsequence of a given 1-2 string */
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= 4; j++) {
dp[j][i] = std::max(dp[j - 1][i], dp[j][i - 1] + (int)(a[i] == s[j]));
}
}
printf("%d\n", dp[4][n]);
return 0;
}