LA3516 Exploring Pyramids

Exploring Pyramids

题目大意：给定一个欧拉序列（即每经过一个点，把这个点加入序列），问有多少种对应的多叉树

序列与树构造对应问题，考虑区间DP

dp[i][j]表示序列i...j对应二叉树个数

初始i == j，dp[i][j] = 1

dp[i][j] = 0,i!=j

转移：dp[i][j] = sum(dp[i + 1][k - 1] * dp[k][j]),s[i] == s[j]  即考虑从i出发，在k这个位置回来，然后再从k出发，到j的时候回来

被MOD卡了一下

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <cmath>
 #define min(a, b) ((a) < (b) ? (a) : (b))
 #define max(a, b) ((a) > (b) ? (a) : (b))
 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
 inline void swap(long long &a, long long &b)
 {
     long long tmp = a;a = b;b = tmp;
 }
 inline void read(long long &x)
 {
     x = ;char ch = getchar(), c = ch;
     while(ch < '' || ch > '') c = ch, ch = getchar();
     while(ch <= '' && ch >= '') x = x *  + ch - '', ch = getchar();
     if(c == '-') x = -x;
 }

 const long long INF = 0x3f3f3f3f;
 const long long MAXN =  + ;
 const long long MOD = ;

 long long dp[MAXN][MAXN],n;
 char s[MAXN];

 int main()
 {
     while(scanf("%s", s + ) != EOF)
     {
         n = strlen(s + );
         memset(dp, , sizeof(dp));
         for(register long long i = ;i <= n;++ i) dp[i][i] = ;
         for(register long long k = ;k <= n;++ k)
             for(register long long i = ;i <= n;++ i)
             {
                 long long j = i + k - ;
                 if(j > n) break;
                 if(s[i] != s[j]) continue;
                 for(register long long k = i + ;k <= j;++ k)
                     if(s[i] == s[k]) dp[i][j] += dp[i + ][k - ] * dp[k][j] % MOD, dp[i][j] = dp[i][j] >= MOD ? dp[i][j] - MOD : dp[i][j];
             }
         printf("%lld\n", dp[][n]);
     }
     return ;
 } 

LA3516