区间dp 注意len==2的特判问题
1 #include<iostream>
2 #include<algorithm>
3 #include<cmath>
4 #include<cstdio>
5 #define INF 1e9
6 using namespace std;
7 const int maxn = 5e2 + 10;
8 int n;
9 int a[maxn];
10 int dp[maxn][maxn];
11
12 int main(){
13 scanf("%d",&n);
14 for(int i = 1 ; i <= n ; i++){
15 for(int j = i ; j <= n ; j++){
16 dp[i][j] = INF;
17 }
18 }
19 for(int i = 1 ; i <= n ; i++){
20 scanf("%d",&a[i]);
21 dp[i][i] = 1;
22 }
23 for(int i = 1 ; i < n ; i++){
24 dp[i][i + 1] = a[i] == a[i + 1] ? 1 : 2;
25 }//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
26 for(int len = 3 ; len <= n ; len++){
27 for(int i = 1 ; i + len - 1 <= n ; i++){
28 int j = i + len - 1;
29 for(int k = i ; k <= j - 1 ; k++){
30 dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
31 }
32 if(a[i] == a[j] && i + 1 <= j - 1) dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);
33 }
34 }
35 printf("%d
",dp[1][n]);
36
37 return 0;
38 }