POJ 2255 Tree Recovery

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<functional>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

#define N 31
#define INF 1000000009
#define eps 0.00000001
#define sf(a) scanf("%d",&a)
const int MAXN = 1e5 + 3;

char pre[N], mid[N];
int L[MAXN], R[MAXN];
int build(int l1, int r1, int l2, int r2)//l1 r1 是在前序序列的范围 l2 r2 是中序的范围
{
    if (l1 > r1 || l2 > r2)
        return -1;
    if (l1 == r1&&l2 == r2)
        return l1;
    char tr = pre[l1];
    int p = l2;
    for (; p <= r2; p++)
    {
        if (mid[p] == tr)
            break;
    }
    L[l1] = build(l1 + 1, l1 + p - l2, l2, p - 1);
    R[l1] = build(l1 + p - l2 + 1, r1, p + 1, r2);
    return l1;
}
void print(int p)
{
    if (L[p] != -1)
        print(L[p]);
    if (R[p] != -1)
        print(R[p]);
    printf("%c", pre[p]);
}
int main()
{
    while (scanf("%s%s", pre, mid) != EOF)
    {
        memset(L, -1, sizeof(L));
        memset(R, -1, sizeof(R));
        int l = strlen(pre);
        int root = build(0, l - 1, 0, l - 1);
        print(root);
        printf("\n");
    }
}