【区间DP/字符串】kiminonamaewa 校OJ2475

//  O(n^4)

#include <bits/stdc++.h>
using namespace std;
 
const int mod = 1e9 + 7;
 
char ch[60];
int sum[60][60];
 
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int len;
        scanf("%d", &len);
        scanf("%s", ch + 1);
 
        memset(sum, 0, sizeof sum);
        for (int i = 1; i <= len; i++)
            sum[i][i] = 1;
 
        //以i j 为两端点
        for (int l = 1; l <= len; l++) //区间长度
        {
            for (int i = 1; i <= len; i++) //起点
            {
                int j = i + l - 1;
                if(j > len) continue;
                if (ch[i] == ch[j])
                {
                    sum[i][j] = 1;
                    for (int u = i + 1; u < j; u++) //枚举中间各起终点情况
                    {
                        for (int v = u; v < j; v++)
                        {
                            sum[i][j] = (sum[i][j] + sum[u][v]) % mod;
 
                        }
                    }
                }
                //printf("[%d, %d] = %d\n", i, j, sum[i][j]);
            }
        }
 
        int ans = 0;
        for (int i = 1; i <= len; i++)  //各起终点回文串数量总和
        {
            for (int j = i; j <= len; j++)
            {
                ans = (ans + sum[i][j]) % mod;
            }
        }
 
        printf("%d\n", ans);
    }
 
    return 0;
}

//  O(n^2)

#include <bits/stdc++.h>
using namespace std;
 
const int mod = 1e9 + 7;
 
char ch[60];
int sum[60][60];
int f[60][60];  //以ij为区间
 
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int len;
        scanf("%d", &len);
        scanf("%s", ch + 1);
 
        memset(sum, 0, sizeof sum);
        memset(f, 0, sizeof f);
         
 
        //以i j 为区间
        for (int l = 1; l <= len; l++) //区间长度
        {
            for (int i = 1; i <= len; i++) //左区间
            {
                int j = i + l - 1;
                if(j > len) continue;
                
                
                // 一.容斥原理 1.sum数组记以ij为端点的回文串数
                if (ch[i] == ch[j])
                    sum[i][j] = (1 + f[i + 1][j - 1]) % mod;
					//容斥原理 关系式
                f[i][j] = (f[i][j - 1] + f[i + 1][j]) % mod;
                f[i][j] = (f[i][j] - f[i + 1][j - 1] + mod) % mod; 
                f[i][j] = (f[i][j] + sum[i][j]) % mod;
                
                // 一.容斥原理 2.sum[i][j]数组用临时变量x代替
				int x = 0;
                if (ch[i] == ch[j]) 
                    x = (1 + f[i + 1][j - 1]) % mod;
                f[i][j] = (f[i][j - 1] + f[i + 1][j]) % mod;
                f[i][j] = (f[i][j] - f[i + 1][j - 1] + mod) % mod; 
                f[i][j] = (f[i][j] + x) % mod;
                
                // 二.直接递推
				f[i][j] = (f[i][j - 1] + f[i + 1][j]) % mod;
                f[i][j] = (f[i][j] - f[i + 1][j - 1] + mod) % mod;
                if (ch[i] == ch[j])
                    f[i][j] = (f[i][j] + f[i + 1][j - 1] + 1) % mod;
            }
        }
 
        printf("%d\n", f[1][len]);
    }
 
    return 0;
}