xtu1143 Data Encoding（huffman）

Data Encoding
Accepted : 14   Submit : 118
Time Limit : 1000 MS   Memory Limit : 1048576 KB

 

Data Encoding

Given a string S, your job is to encode it into a new string T using given charset C. To make the decoding easy, each original character X in S is encoded into a string Y made up of characters in C. What's more, for two different X1 and X2, Y1 must be different from Y2 and all its non-empty prefixes. What's the minimum possible length of T?

Input

There are multiple test cases. Each test case contains two lines, S and C. The length of S will be at most 100000. The length of C will be at least 2 and all characters in C will be different.

Output

For each test case, output the minimum possible length of T.

Sample Input

test
01
1267650600228229401496703205376
ATCG

Sample Output

6
50

题意：给定你一个原串s1，现在要用s2编码。

分析：如果s2的长度为2就是就是经典的huffman编码了，这里其实也一样，由原来的2叉变为k叉。

注意：输入我cin读入字符串 WA，后改用gets AC了，中间可能会有空格。

         还有一点就是，有可能叶子节点不能满k叉，所以要加以注意。

View Code 
 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstring>
 4 #define N 100010
 5 
 6 using namespace std;
 7 
 8 char s1[N],s2[N];
 9 int num[N];
10 
11 int huffman(int n,int k)
12 {
13     int i,j,sum,ans=0;
14     if(n<=k) 
15     {
16         for(i=0;i<n;i++)
17             ans+=num[i];
18         return ans;
19     }
20     int m=(n-1)%(k-1);
21     if(m)
22     {
23         sum=0;
24         for(i=0;i<=m;i++)
25         {
26             sum+=num[i];
27         }
28         ans+=sum;
29         for(i=m+1;i<n;i++)
30         {
31             if(sum>num[i])
32                 num[i-1]=num[i];
33             else break;
34         }
35         num[i-1]=sum;
36     }
37     for(i=m;i<n-1;i=i+k-1)
38     {
39         sum=0;
40         for(j=i;j<i+k&&j<n;j++)
41         {
42             sum+=num[j];
43         }
44         ans+=sum;
45         for(j=i+k;j<n;j++)
46         {
47             if(sum>num[j])
48                 num[j-1]=num[j];
49             else break;
50         }
51         num[j-1]=sum;
52     }
53     return ans;
54 }
55 
56 int main()
57 {
58     int i,k,len1,len2;
59     while(getline(s1))//gets()读一行 cin输入WA 错误不明！！
60     {
61         getline(s2);
62         k=strlen(s2);
63         len2=strlen(s1);
64         sort(s1,s1+len2);
65         len1=0;
66         for(i=0;i<len2;i++)
67         {
68             num[len1]=1;
69             while(s1[i]==s1[i+1])
70             {
71                 num[len1]++;
72                 i++;
73             }
74             len1++;
75         }
76         sort(num,num+len1);
77         cout<<huffman(len1,k)<<endl;
78     }
79     return 0;
80 }
81 /*
82 test
83 01
84 
85 1267650600228229401496703205376
86 ATCG
87 
88 12233444
89 123
90 
91 123456 123
92 
93 */

 

posted @ 2012-05-20 08:36  mtry  阅读(230)  评论(0)    收藏  举报