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 = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}

const long long INF = 0x3f3f3f3f;
const long long MAXN = 300 + 10;
const long long MOD = 1000000000;

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

int main()
{
    while(scanf("%s", s + 1) != EOF)
    {
        n = strlen(s + 1);
        memset(dp, 0, sizeof(dp));
        for(register long long i = 1;i <= n;++ i) dp[i][i] = 1;
        for(register long long k = 1;k <= n;++ k)
            for(register long long i = 1;i <= n;++ i)
            {
                long long j = i + k - 1;
                if(j > n) break;
                if(s[i] != s[j]) continue;
                for(register long long k = i + 2;k <= j;++ k)
                    if(s[i] == s[k]) dp[i][j] += dp[i + 1][k - 1] * dp[k][j] % MOD, dp[i][j] = dp[i][j] >= MOD ? dp[i][j] - MOD : dp[i][j];
            }
        printf("%lld\n", dp[1][n]);
    }
    return 0;
}