Longest Palindrome - UVa 11151 dp

Problem D: Longest Palindrome

Time limit: 10 seconds

A palindrome is a string that reads the same from the left as it does from the right. For example, I, GAG and MADAM are palindromes, but ADAM is not. Here, we consider also the empty string as a palindrome.

From any non-palindromic string, you can always take away some letters, and get a palindromic subsequence. For example, given the string ADAM, you remove the letter M and get a palindrome ADA.

Write a program to determine the length of the longest palindrome you can get from a string.

Input and Output

The first line of input contains an integer T (≤ 60). Each of the next T lines is a string, whose length is always less than 1000.

For ≥90% of the test cases, string length ≤ 255.

For each input string, your program should print the length of the longest palindrome you can get by removing zero or more characters from it.

Sample Input

2ADAMMADAM

Sample Output

35

题意：最长非连续回文字串。

思路：dp[i][j]表示从i到j的最长回文字串长度。注意测试数据有坑，有空行，还有 scanf("%d",&t);后面要加getchar()，不能用scanf("%d\n",&t);因为t后面可能有空格后再回车…………WA了好多发……真丢人。

AC代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[1010][1010];char s[1010];int main(){ int t,n,i,j,k;  scanf("%d",&t);  getchar();  while(t--)  { gets(s);    n=strlen(s);    memset(dp,0,sizeof(dp));    for(i=0;i<n;i++)     dp[i][i]=1;    for(i=n-2;i>=0;i--)     for(j=i+1;j<n;j++)     { dp[i][j]=max(dp[i+1][j],dp[i][j-1]);       if(s[i]==s[j])        dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);     }    printf("%d\n",dp[0][n-1]);  }}