C. A Twisty Movement
time limit per test1 second
memory limit per test256 megabytes
Problem Description
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.
Input
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).
Output
Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.
Examples
input
4
1 2 1 2
output
4
input
10
1 1 2 2 2 1 1 2 2 1
output
9
Note
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.
解题心得：
- 题意很简单就是输出最长不递减子序列的长度。
- 读错题了啊，把子序列看成子区间弄错了，。
- 就是一个dp加一点思维
- 先求出1的前缀和，2的后缀和
- dp[i][j][1]代表在区间（i，j）之间以1结尾的最长不递增子序列的长度。
dp[i][j][2]代表在区间（i，j）之间以2结尾的最长不递增子系列的长度。 - 状态转移方程式就很容易出来了dp[i][j][1] = max(dp[i][j-1][1],dp[i][j-1][2]) + (num[j] == 1)
dp[i][j][2] = dp[i][j-1][2] + （num[j] == 2）
- 至于为什么要求不递增的dp，那就是要翻转啊，翻转之后不递增不就变成不递减了吗。而前缀和和后缀和就是考考思维。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2010;
int dp[maxn][maxn][3],num[maxn],sum1[maxn],sum2[maxn];
int Max = -1,n;
int main(){
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
for(int i=0;i<n;i++)
sum1[i] = sum1[i-1] + (num[i] == 1);
for(int i=n-1;i>=0;i--)
sum2[i] = sum2[i+1] + (num[i] == 2);
for(int i=0;i<n;i++)
for(int j=i;j<n;j++){
dp[i][j][2] = dp[i][j-1][2] + (num[j] == 2);
dp[i][j][1] = max(dp[i][j-1][1],dp[i][j-1][2]) + (num[j] == 1);
Max = max(dp[i][j][1] + sum1[i-1] + sum2[j+1],Max);
Max = max(dp[i][j][2] + sum1[i-1] + sum2[j+1],Max);
}
printf("%d",Max);
return 0;
}