Codeforces 245H 回文 DP

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本文详细介绍了如何通过状态转移方程来计算字符串中回文子串的数量，包括初始化、回文子串判断及动态规划求解过程。此方法为解决字符串处理问题提供了一种高效策略。

看了解题报告才弄清楚状态转移方程：

Let isPal[i][j] be 1 if s[i...j] is palindrome, otherwise, set it 0. Let's define dp[i][j] to be number of palindrome substrings of s[i...j]. Let's calculate isPal[i][j] and dp[i][j] in O(|S|2). First, initialize isPal[i][i] = 1 and dp[i][i] = 1. After that, loop over len which states length of substring and for each specific len, loop over start which states starting position of substring. isPal[start][start + len - 1]can be easily calculated by the following formula:

isPal[start][start+len-1] = isPal[start+1][start+len-2] & (s[start] == s[start+len-1])

After that, dp[start][start + len - 1] can be calculated by the following formula which is derived from Inc-Exc Principle.

dp[start][start+len-1] = dp[start][start+len-2] + dp[start+1][start+len-1] - dp[start+1][start+len-2] + isPal[start][start+len-1]

#include<cstdio>
#include<iostream>
#include<cstring>
#define MAXN 5001
using namespace std;
int f[MAXN][MAXN],d[MAXN][MAXN];
int main()
{
   // freopen("in.txt","r",stdin);
    char str[MAXN];
    cin>>str;
    int len=strlen(str);
    memset(f,0,sizeof(f));
    memset(d,0,sizeof(d));
    for(int i=len-1;i>=0;i--)
    {
        f[i][i]=1;
        f[i][i-1]=1;
        for(int j=i+1;j<len;j++)
        {
            if(str[j]==str[i]&&f[i+1][j-1])
                f[i][j]=1;
        }
    }
    for(int i=len-1;i>=0;i--)
    {
        d[i][i]=1;
        for(int j=i+1;j<len;j++)
        {
            d[i][j]=d[i+1][j]+d[i][j-1]-d[i+1][j-1];
            if(f[i][j])
                d[i][j]++;
        }
    }
    int n,a,b;
    cin>>n;
    while(n--)
    {
        cin>>a>>b;
        cout<<d[a-1][b-1]<<endl;
    }
    return 0;
}