L2-008 最长对称子串

题解

• vis[i][j]表示[i,j]是否为对称子串，dp[i][j]表示[i,j]最大的对称子串的长度
• if (s[i] == s[j] && vis[i+1][j-1])  dp[i][j] = dp[i+1][j-1] + 2，vis[i][j] = true;
• else dp[i][j] = max(max(dp[i+1][j],dp[i][j-1]),dp[i+1][j-1]);

题意

[code]#include <bits/stdc++.h>
using namespace std;
int const N = 1000 + 10;
int dp

,vis

;
char s
;
int main()
{
cin.getline(s,N);
int len = strlen(s);
for(int i=0;i<len;i++){
dp[i][i] = 1;
vis[i][i] = true;
if(i != len - 1 && s[i+1] == s[i]){
dp[i][i+1] = 2;
vis[i][i+1] = true;
}
}
for(int k=3;k<=len;k++)
for(int i=0;i+k-1<len;i++){
int j = i + k - 1;
if(s[i] == s[j] && vis[i+1][j-1] == true){
vis[i][j] = true;
dp[i][j] = dp[i+1][j-1] + 2;
}else   dp[i][j] = max(max(dp[i+1][j],dp[i][j-1]),dp[i+1][j-1]);
}
cout<<dp[0][len-1]<<endl;
return 0;
}