51nod1832(二叉树/高精度模板+dfs)

题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1832

题意: 中文题诶~

思路: 若二叉树中有 k 个节点只有一个子树, 则答案为 1 << k.

详情参见:http://blog.csdn.net/gyhguoge01234/article/details/77836484

代码:

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #define ll long long
  5 using namespace std;
  6 
  7 const int MAX = 1e2;
  8 const int M = 1e9;//1e9为1节
  9 const int MAXN = 35;
 10 
 11 struct BigInt{
 12     const static int mod = 10000;
 13     const static int DLEN = 4;
 14     int a[600], len;
 15     BigInt(){
 16         memset(a, 0, sizeof(a));
 17         len = 1;
 18     }
 19     BigInt(int v){
 20         memset(a, 0, sizeof(a));
 21         len = 0;
 22         do{
 23             a[len++] = v % mod;
 24             v /= mod;
 25         }while(v);
 26     }
 27     BigInt(const char s[]){
 28         memset(a, 0, sizeof(a));
 29         int L = strlen(s);
 30         len = L / DLEN;
 31         if(L % DLEN) len++;
 32         int index = 0;
 33         for(int i = L - 1; i >= 0; i -= DLEN){
 34             int t = 0;
 35             int k = i - DLEN + 1;
 36             if(k < 0) k = 0;
 37             for(int j = k; j <= i; j++)
 38                 t = t * 10 + s[j] - '0';
 39             a[index++] = t;
 40         }
 41     }
 42     BigInt operator +(const BigInt &b)const{
 43         BigInt res;
 44         res.len = max(len, b.len);
 45         for(int i = 0; i <= res.len; i++) res.a[i] = 0;
 46         for(int i = 0; i < res.len; i++){
 47             res.a[i] += ((i < len) ? a[i] : 0) + ((i < b.len) ? b.a[i] : 0);
 48             res.a[i + 1] += res.a[i] / mod;
 49             res.a[i] %= mod;
 50         }
 51         if(res.a[res.len] > 0) res.len++;
 52         return res;
 53     }
 54     BigInt operator *(const BigInt &b)const{
 55         BigInt res;
 56         for(int i = 0; i < len; i++){
 57             int up = 0;
 58             for(int j = 0; j < b.len; j++){
 59                 int temp = a[i] * b.a[j] + res.a[ i + j] + up;
 60                 res.a[i + j] = temp%mod;
 61                 up = temp / mod;
 62             }
 63             if(up != 0)
 64             res.a[i + b.len] = up;
 65         }
 66         res.len = len + b.len;
 67         while(res.a[res.len - 1] == 0 && res.len > 1) res.len--;
 68         return res;
 69     }
 70     void output(){
 71         printf("%d", a[len - 1]);
 72         for(int i = len - 2; i >= 0; i--)
 73             printf("%04d", a[i]);
 74         printf("\n");
 75     }
 76 };
 77 
 78 // 先序遍历 X L … R …
 79 // 后序遍历 … L … R X
 80 
 81 const int N = 1e5 + 10;
 82 int a[N], b[N];
 83 BigInt sol(1);
 84 
 85 void dfs(int al, int ar, int bl, int br){
 86     if(ar - al <= 1) return;
 87     al++;
 88     br--;
 89     int indx = bl, cnt = 0;;
 90     while(a[al] != b[indx]) indx++;
 91     int newar = al + (indx - bl + 1);
 92     int newbr = indx + 1;
 93     cnt++;
 94     dfs(al, newar, bl, newbr);
 95     if(ar - al != indx - bl + 1){
 96         cnt++;
 97         dfs(newar, ar, newbr, br);
 98     }
 99     if(cnt == 1) sol = sol * 2;
100 }
101 
102 int main(void){
103     int n;
104     scanf("%d", &n);
105     for(int i = 0; i < n; i++){
106         scanf("%d", &a[i]);
107     }
108     for(int i = 0; i < n; i++){
109         scanf("%d", &b[i]);
110     }
111     sol = 1;
112     dfs(0, n, 0, n);
113     sol.output();
114     return 0;
115 }