codeforces 1132F 区间DP

codeforces 1132F

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题意：

$给定一串长度为n的字符，你可以以任意顺序删除所有字母相同的子串。$
$问删除所有字符最少操作数。$

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题解：

$dp[l][r]表示删除区间[l,r]内所有字符的最优解。$

• $s[l]=s[r]时，dp[l][r]=dp[l+1][r-1]+1$
• $s[l]̸=s[r]，dp[l][r]=min(dp[l+1][r],dp[l][r-1])+1$
• $枚举分割点k，dp[l][r]=min(dp[l][r],dp[l][k]+dp[k][r]-1)$

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#include 
using namespace std;
const int N = 501;
char s[N];
int dp[N][N];

int main() {
    int n;
    cin >> n;
    for(int i = 1 ; i <= n ; i++){
        cin >> s[i];
    }
    for(int i = 1 ; i <= n ; i++){
        dp[i][i] = 1;
    }
    for(int len = 2 ; len <= n ; len++){
        for(int l = 1, r = len ; r <= n ; l++, r++){
            if(s[l] == s[r]){
                dp[l][r] = dp[l+1][r-1]+1;
            }
            else{
                dp[l][r] = min(dp[l+1][r], dp[l][r-1])+1;
            }
            for(int k = l ; k <= r ; k++){
                dp[l][r] = min(dp[l][r], dp[l][k]+dp[k][r]-1);
            }
        }
    }
    cout << dp[1][n] << endl;
    return 0; 
}