poj 3464(Trie)Approximations

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Approximations

Time Limit: 2000MS		Memory Limit: 131072K
Total Submissions: 419		Accepted: 23

Description

For any decimal fraction, we can obtain a set of approximations of different accuracy by mean of rounding. Take 0.2503 for example, we have the following approximations:

• 0.2503
• 0.250
• 0.25
• 0.3
• 0.

If two fractions A and B can both be rounded to C, we call C a common approximation of A and B. Two fractions may have more than one common approximations, each having a distinct accuracy. For example, 0.2503 and 0.2504 have common approximations 0.250 and 0.25. The accuracy of the former is 10?3, while that of the latter is 10?2. Among all common approximations of two fractions, there is one that has the highest accuracy, and we call it the most accurate common approximation (MACA) of the two fractions. By this definition, the MACA of 0.2503 and 0.2504 is 0.250.

Given N fractions Ai (1 ≤ i ≤ N) in the range [0, 0.5), find a fraction x that maximizes the sum of ?log10 (the accuracy of the MACA of Ai and x). Report that maximized sum.

Input

The first line contains one integer N. N ≤ 100000.
Each of the next N lines contains a decimal fraction Ai. The total number of digits of the N decimal fractions doesn‘t exceed 400000. There is always a radix point, so zero is "0." instead of "0".

Output

One integer, the maximized sum.

Sample Input

40.2500.25060.251150.2597

Sample Output

11

Hint

x = 0.25115.

Source

POJ Monthly--2007.11.25, Yang Yi 

董華星在他09年的論文裡說可以用字典樹寫。。我試著寫了下，然而感覺題目是不是給的範圍有問題啊。測了好多範例沒問題。歡迎各位大佬給個範例測測0 0。

代碼如下：

  1 #include<cstdio>  2 #include<iostream>  3 #include<cstring>  4 #define clr(x) memset(x,0,sizeof(x))  5 #define clr_1(x) memset(x,-1,sizeof(x))  6 #define LL long long  7 #define mod 1000000007  8 #define INF 0x3f3f3f3f  9 #define next nexted 10 using namespace std; 11 const int N=1e5+10; 12 const int M=4e5+10; 13 int next[M][10]; 14 int num[M]; 15 char s[M]; 16 int n,m,k; 17 int root,ttot; 18 void tadd(char *s,int root) 19 { 20     int now=root,p; 21     int len=strlen(s); 22     for(int i=0;i<len;i++) 23     { 24         p=s[i]-‘0‘; 25         if(next[now][p]==0) 26         { 27             next[now][p]=++ttot; 28         } 29         now=next[now][p]; 30         num[now]++; 31     } 32     return ; 33 } 34 int ans,prenode; 35 void dfs(int now,int prenum) 36 { 37     int nownum,p; 38     for(int i=0;i<10;i++) 39     { 40         nownum=prenum; 41         p=next[now][i]; 42         if(i==0) 43         { 44             if(prenode!=0) 45             { 46                 prenode=next[prenode][9]; 47                 if(prenode!=0) 48                 { 49                     for(int j=5;j<10;j++) 50                         if(next[prenode][j]!=0) 51                             nownum+=num[next[prenode][j]]; 52                 } 53             } 54         } 55         else 56         { 57             if(next[now][i-1]!=0) 58                for(int j=5;j<10;j++) 59                 { 60                     if(next[next[now][i-1]][j]!=0) 61                     { 62                         nownum+=num[next[next[now][i-1]][j]]; 63                         if(i-1>=5) 64                         { 65                             nownum+=num[next[next[now][i-1]][j]]; 66                         } 67                     } 68  69                 } 70         } 71         if(i!=0) 72         { 73             prenode=next[now][i-1]; 74         } 75         if(p!=0) 76         { 77             nownum+=num[p]; 78             for(int j=5;j<10;j++) 79             { 80                 if(next[p][j]!=0) 81                 { 82                     nownum-=num[next[p][j]]; 83                 } 84             } 85             if(i>=5) 86                 nownum+=num[p]; 87             if(nownum>ans) 88                 ans=nownum; 89 //            cout<<i<<" "<<nownum<<endl; 90             dfs(p,nownum); 91         } 92         else 93         { 94             if(nownum>ans) 95                 ans=nownum; 96         } 97     } 98     return ; 99 }100 int main()101 {102     scanf("%d",&n);103     ttot=root=0;104     for(int i=1;i<=n;i++)105     {106         scanf("%s",s);107         tadd(s+2,root);108     }109     ans=0;110     dfs(root,0);111     printf("%d\n",ans);112     return 0;113 }

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poj 3464(Trie)Approximations