1040 Longest Symmetric String (25)（25 分）

回文串动态规划：
s[i] == s[j]：dp[i][j]=dp[i+1][j-1]
s[i]!=s[j]:dp[i][j]=0
边界条件：dp[i][i]=1 dp[i][i+1]=(s[i]==s[i+1])?1:0

注意ans初始值为1

#include
#include
#include
using namespace std;
const int maxn = 1e3 + 10;
int dp[maxn][maxn];
int main()
{
    string s;
    getline(cin, s);
    int len = s.length();
    int ans = 1;
    for (int i = 0; i < len; i++)
    {
        dp[i][i] = 1;
        if (i < len - 1 && s[i] == s[i + 1])
        {
            dp[i][i + 1] = 1;
            ans = 2;
        }
    }
    for (int L = 3; L <= len; L++)
    {
        for (int i = 0; i + L - 1 < len; i++)
        {
            int j = i + L - 1;
            if (s[i] == s[j] && dp[i + 1][j - 1] == 1)
            {
                dp[i][j] = 1;
                ans = L;
            }
        }
    }
    printf("%d", ans);
    return 0;
}