CF607B

区间dp 注意len==2的特判问题

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define INF 1e9
using namespace std;
const int maxn = 5e2 + 10;
int n;
int a[maxn];
int dp[maxn][maxn];

int main(){
    scanf("%d",&n);
    for(int i = 1 ; i <= n ; i++){
        for(int j = i ; j <= n ; j++){
            dp[i][j] = INF;
        }
    }
    for(int i = 1 ; i <= n ; i++){
        scanf("%d",&a[i]);
        dp[i][i] = 1;
    }
    for(int i = 1 ; i < n ; i++){
        dp[i][i + 1] = a[i] == a[i + 1] ? 1 : 2;
    }//若枚举len从2开始，dp[i+1][j-1]将无法被覆盖，因此需要特判 
    for(int len = 3 ; len <= n ; len++){
        for(int i = 1 ; i + len - 1 <= n ; i++){
            int j = i + len - 1;
            for(int k = i ; k <= j - 1 ; k++){
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
            }
            if(a[i] == a[j] && i + 1 <= j - 1)    dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);
        }
    }
    printf("%d
",dp[1][n]);
    
    return 0;
}