HDU 4691 Front compression （2013多校9 1006题 后缀数组）

Front compression

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 158    Accepted Submission(s): 63


Problem Description
Front compression is a type of delta encoding compression algorithm whereby common prefixes and their lengths are recorded so that they need not be duplicated. For example:

The size of the input is 43 bytes, while the size of the compressed output is 40. Here, every space and newline is also counted as 1 byte.
Given the input, each line of which is a substring of a long string, what are sizes of it and corresponding compressed output?
 

 

Input
There are multiple test cases. Process to the End of File.
The first line of each test case is a long string S made up of lowercase letters, whose length doesn't exceed 100,000. The second line contains a integer 1 ≤ N ≤ 100,000, which is the number of lines in the input. Each of the following N lines contains two integers 0 ≤ A < B ≤ length(S), indicating that that line of the input is substring [A, B) of S.
 

 

Output
For each test case, output the sizes of the input and corresponding compressed output.
 

 

Sample Input
frcode 2 0 6 0 6 unitedstatesofamerica 3 0 6 0 12 0 21 myxophytamyxopodnabnabbednabbingnabit 6 0 9 9 16 16 19 19 25 25 32 32 37
 

 

Sample Output
14 12 42 31 43 40
 

 

Author
Zejun Wu (watashi)
 

 

Source
2013 Multi-University Training Contest 9
 

 

Recommend
zhuyuanchen520
 

 

 

 

后缀数组随便搞一下就可以了

 

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013/8/20 13:40:03
  4 File Name     :F:\2013ACM练习\2013多校9\1006.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MAXN=100010;
 21 int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
 22 //待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
 23 //除s[n-1]外的所有s[i]都大于0，r[n-1]=0
 24 //函数结束以后结果放在sa数组中
 25 bool cmp(int *r,int a,int b,int l)
 26 {
 27     return r[a] == r[b] && r[a+l] == r[b+l];
 28 }
 29 void da(int str[],int sa[],int rank[],int height[],int n,int m)
 30 {
 31     n++;
 32     int i, j, p, *x = t1, *y = t2;
 33     //第一轮基数排序，如果s的最大值很大，可改为快速排序
 34     for(i = 0;i < m;i++)c[i] = 0;
 35     for(i = 0;i < n;i++)c[x[i] = str[i]]++;
 36     for(i = 1;i < m;i++)c[i] += c[i-1];
 37     for(i = n-1;i >= 0;i--)sa[--c[x[i]]] = i;
 38     for(j = 1;j <= n; j <<= 1)
 39     {
 40         p = 0;
 41         //直接利用sa数组排序第二关键字
 42         for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小
 43         for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
 44         //这样数组y保存的就是按照第二关键字排序的结果
 45         //基数排序第一关键字
 46         for(i = 0; i < m; i++)c[i] = 0;
 47         for(i = 0; i < n; i++)c[x[y[i]]]++;
 48         for(i = 1; i < m;i++)c[i] += c[i-1];
 49         for(i = n-1; i >= 0;i--)sa[--c[x[y[i]]]] = y[i];
 50         //根据sa和x数组计算新的x数组
 51         swap(x,y);
 52         p = 1; x[sa[0]] = 0;
 53         for(i = 1;i < n;i++)
 54             x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
 55         if(p >= n)break;
 56         m = p;//下次基数排序的最大值
 57     }
 58     int k = 0;
 59     n--;
 60     for(i = 0;i <= n;i++)rank[sa[i]] = i;
 61     for(i = 0;i < n;i++)
 62     {
 63         if(k)k--;
 64         j = sa[rank[i]-1];
 65         while(str[i+k] == str[j+k])k++;
 66         height[rank[i]] = k;
 67     }
 68 }
 69 int rank[MAXN],height[MAXN];
 70 int RMQ[MAXN];
 71 int mm[MAXN];
 72 int best[20][MAXN];
 73 void initRMQ(int n)
 74 {
 75     mm[0]=-1;
 76     for(int i=1;i<=n;i++)
 77         mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
 78     for(int i=1;i<=n;i++)best[0][i]=i;
 79     for(int i=1;i<=mm[n];i++)
 80         for(int j=1;j+(1<<i)-1<=n;j++)
 81         {
 82             int a=best[i-1][j];
 83             int b=best[i-1][j+(1<<(i-1))];
 84             if(RMQ[a]<RMQ[b])best[i][j]=a;
 85             else best[i][j]=b;
 86         }
 87 }
 88 int askRMQ(int a,int b)
 89 {
 90     int t;
 91     t=mm[b-a+1];
 92     b-=(1<<t)-1;
 93     a=best[t][a];b=best[t][b];
 94     return RMQ[a]<RMQ[b]?a:b;
 95 }
 96 int lcp(int a,int b)
 97 {
 98     a=rank[a];b=rank[b];
 99     if(a>b)swap(a,b);
100     return height[askRMQ(a+1,b)];
101 }
102 char str[MAXN];
103 int r[MAXN];
104 int sa[MAXN];
105 int A[MAXN],B[MAXN];
106 int calc(int n)
107 {
108     if(n == 0)return 1;
109     int ret = 0;
110     while(n)
111     {
112         ret++;
113         n /= 10;
114     }
115     return ret;
116 }
117 int main()
118 {
119     //freopen("in.txt","r",stdin);
120     //freopen("out.txt","w",stdout);
121     while(scanf("%s",str)==1)
122     {
123         int n = strlen(str);
124         for(int i = 0;i < n;i++)
125             r[i] = str[i];
126         r[n] = 0;
127         da(r,sa,rank,height,n,128);
128         for(int i = 1;i <= n;i++)
129             RMQ[i] = height[i];
130         initRMQ(n);
131         int k,u,v;
132         long long ans1 = 0, ans2 = 0;
133         scanf("%d",&k);
134         for(int i = 0;i < k;i++)
135         {
136             scanf("%d%d",&A[i],&B[i]);
137             if(i == 0)
138             {
139                 ans1 += B[i] - A[i] + 1;
140                 ans2 += B[i] - A[i] + 3;
141                 continue;
142             }
143             int tmp ;
144             if(A[i]!= A[i-1])tmp = lcp(A[i],A[i-1]);
145             else tmp = 10000000;
146             tmp = min(tmp,B[i]-A[i]);
147             tmp = min(tmp,B[i-1]-A[i-1]);
148             ans1 += B[i] - A[i] + 1;
149             ans2 += B[i] - A[i] - tmp + 1;
150             ans2 += 1;
151             ans2 += calc(tmp);
152         }
153         printf("%I64d %I64d\n",ans1,ans2);
154     }
155     return 0;
156 }

 

 

 

posted on 2013-08-20 19:32  kuangbin  阅读(1786)  评论(0)  编辑  收藏  举报

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