标签:ges scan div hid 没有 logs one 字符串 树遍历
题意:给定一个字符串,求有多少种树与之对应,对应方式是,每次遍历左节点,没有了,就回溯;
分析:d[i,j] = sum(d[i+1,k-1],d[k,j]) (str[i]==str[k]);
坑点是数组竟然要long long 不然会超时,神奇;
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 300+5; 6 const int mod = 1000000000; 7 char str[maxn]; 8 int d[maxn][maxn]; 9 10 11 int dp(int i,int j) { 12 if(i==j) return 1; 13 if(str[i]!=str[j]) return 0; 14 int& ans = d[i][j]; 15 if(ans>=0) return ans; 16 ans = 0; 17 18 for(int k=i+2;k<=j;k++) { 19 if(str[i]==str[k]) { 20 ans = (ans + (long long)dp(i+1,k-1)*(long long)dp(k,j)) % mod; 21 } 22 } 23 return ans; 24 } 25 26 int main() 27 { 28 freopen("exploring.in","r",stdin); 29 freopen("exploring.out","w",stdout); 30 while(scanf("%s",str)!=EOF) { 31 memset(d,-1,sizeof(d)); 32 printf("%d\n",dp(0,strlen(str)-1)); 33 } 34 return 0; 35 }
标签:ges scan div hid 没有 logs one 字符串 树遍历
原文地址:http://www.cnblogs.com/TreeDream/p/6937997.html