Codeforces 245H Queries for Number of Palindromes

http://codeforces.com/contest/245/problem/H

题意:给定一个字符串，每次给个区间，求区间内有几个回文串(n<=5000)

思路:设定pd[i][j]代表i~j这部分是不是一个回文串，这个可以n^2预处理

然后设定f[i][j]代表i~j区间有多少个回文串，由于满足区间加减，因此有

f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1]+pd[i][j]

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
int pd[5005][5005],f[5005][5005],n;
char s[20005];
int read(){
    int t=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
    while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();}
    return t*f;
}
void init(){
    for (int i=1;i<=n;i++)
     pd[i][i]=1;
    for (int len=2;len<=n;len++)
     for (int i=1;i+len-1<=n;i++){
            int j=i+len-1;
            if (i==j-1&&s[i]==s[j]) pd[i][j]=1;
            else
            if (s[i]==s[j]) pd[i][j]|=pd[i+1][j-1];
    }
    for (int i=1;i<=n;i++)
     f[i][i]=1;
    for (int i=1;i<n;i++)
     if (s[i]==s[i+1])
      f[i][i+1]=3;
     else
      f[i][i+1]=2; 
    for (int len=3;len<=n;len++)
     for (int i=1;i+len-1<=n;i++){
        int j=i+len-1;
        f[i][j]=f[i+1][j]+f[i][j-1]-f[i+1][j-1]+pd[i][j];
     }   
}
int main(){
    scanf("%s",s+1);
    n=strlen(s+1);
    init();
    int T=read();
    while (T--){
        int l=read(),r=read();
        printf("%d
",f[l][r]);
    }
    return 0;
}