我的思路是枚举ij为前后端点的区间，然后该区间是由[i,k]和[k+1][j]更新得到。 请问这题一定要严格按照先把所有小区间推完再推大区间的思路做吗，如果按我推的顺序是会有遗漏吗？ 求各位大佬解答…

程序如下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 205
  using namespace std;
char ch1[5][N] , ch2[5][N];
char s[N];
int dp[N][N][5];
int num[N];
int change_num(char ch){
    if(ch == 'W') return 1;
    if(ch == 'I') return 2;
    if(ch == 'N') return 3;
    if(ch == 'G') return 4;
}
char change_ch(int x){
    if(x == 1) return 'W';
    if(x == 2) return 'I';
    if(x == 3) return 'N';
    if(x == 4) return 'G';
}
int main()
{
    for(int i = 1 ; i <= 4 ; i++) cin >> num[i];
    for(int i = 1 ; i <= 4 ; i++)
        for(int j = 1 ; j <= num[i] ; j++) cin >> ch1[i][j] >> ch2[i][j];
    scanf("%s" , s + 1);
    int n = strlen(s + 1);
    for(int i = n - 1 ; i >= 1 ; i -= 2)
        for(int j = i + 1 ; j <= n ; j += 2)
        {
            if(j == i + 1){
                for(int k = 1 ; k <= 4 ; k++)
                    for(int w = 1 ; w <= num[k] ; w++)
                        if(s[i] == ch1[k][w] && s[j] == ch2[k][w]) dp[i][j][k] = 1;
                continue;
            }
            for(int k = i + 1 ; k <= n - 2 ; k += 2){
                for(int p = 1 ; p <= 4 ; p ++)
                    for(int w = 1 ; w <= num[p] ; w++){
                        if(dp[i][k][change_num(ch1[p][w])] && dp[k + 1][j][change_num(ch2[p][w])]) dp[i][j][p] = 1;
                    }
            }
        }
    bool f = 0;
    for(int i = 1 ; i <= 4 ; i++)
        if(dp[1][n][i]){
            cout << change_ch(i);
            f = 1;
        }
    if(!f) cout << "The name is wrong!";
    return 0;
}