HDU 4691 Front compression

Front compression

Time Limit: 5000ms

Memory Limit: 102400KB

This problem will be judged on  HDU. Original ID: 4691
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

2013 Multi-University Training Contest 9

 

解题：后缀数组配合RMQ的使用

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 #include <climits>

  7 #include <vector>

  8 #include <queue>

  9 #include <cstdlib>

 10 #include <string>

 11 #include <set>

 12 #include <stack>

 13 #define LL long long

 14 #define pii pair<int,int>

 15 #define INF 0x3f3f3f3f

 16 using namespace std;

 17 const int maxn = 100010;

 18 int rk[maxn],wb[maxn],wv[maxn],wd[maxn],lcp[maxn];

 19 bool cmp(int *r,int i,int j,int k){

 20     return r[i] == r[j] && r[i+k] == r[j+k];

 21 }

 22 void da(int *r,int *sa,int n,int m){

 23     int i,k,p,*x = rk,*y = wb;

 24     for(i = 0; i < m; ++i) wd[i] = 0;

 25     for(i = 0; i < n; ++i) wd[x[i] = r[i]]++;

 26     for(i = 1; i < m; ++i) wd[i] += wd[i-1];

 27     for(i = n-1; i >= 0; --i) sa[--wd[x[i]]] = i;

 28 

 29     for(p = k = 1; p < n; k <<= 1,m = p){

 30         for(p = 0,i = n-k; i < n; ++i) y[p++] = i;

 31         for(i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;

 32         for(i = 0; i < n; ++i) wv[i] = x[y[i]];

 33 

 34         for(i = 0; i < m; ++i) wd[i] = 0;

 35         for(i = 0; i < n; ++i) wd[wv[i]]++;

 36         for(i = 1; i < m; ++i) wd[i] += wd[i-1];

 37         for(i = n-1; i >= 0; --i) sa[--wd[wv[i]]] = y[i];

 38 

 39         swap(x,y);

 40         x[sa[0]] = 0;

 41         for(p = i = 1; i < n; ++i)

 42             x[sa[i]] = cmp(y,sa[i-1],sa[i],k)?p-1:p++;

 43     }

 44 }

 45 void calcp(int *r,int *sa,int n){

 46     for(int i = 1; i <= n; ++i) rk[sa[i]] = i;

 47     int h = 0;

 48     for(int i = 0; i < n; ++i){

 49         if(h > 0) h--;

 50         for(int j = sa[rk[i]-1]; i+h < n && j+h < n; h++)

 51             if(r[i+h] != r[j+h]) break;

 52         lcp[rk[i]] = h;

 53     }

 54 }

 55 int st[maxn][20];

 56 void init(int n){

 57     memset(st,0,sizeof(st));

 58     for(int i = 1; i <= n; ++i) st[i][0] = lcp[i];

 59     for(int i = 1; 1<<i <= n; ++i){

 60         for(int j = 1; j+(1<<i) <= n+1; ++j)

 61             st[j][i] = min(st[j][i-1],st[j+(1<<(i-1))][i-1]);

 62     }

 63 }

 64 int query(int s,int t){

 65     s = rk[s];

 66     t = rk[t];

 67     if(s > t) swap(s,t);

 68     s++;

 69     int k = log2(t - s + 1);

 70     return min(st[s][k],st[t-(1<<k)+1][k]);

 71 }

 72 int r[maxn],sa[maxn];

 73 char str[maxn];

 74 int mb(int x){

 75     int ans = 0;

 76     if(x == 0) return 1;

 77     while(x){

 78         x /= 10;

 79         ++ans;

 80     }

 81     return ans;

 82 }

 83 int main() {

 84     int hn,x,y;

 85     while(~scanf("%s",str)){

 86         int len = strlen(str);

 87         for(int i = 0; str[i]; ++i)

 88             r[i] = str[i];

 89         r[len] = 0;

 90         da(r,sa,len+1,128);

 91         calcp(r,sa,len);

 92         init(len);

 93         LL ans = 0,ans2 = 0;

 94         int px,py;

 95         scanf("%d",&hn);

 96         for(int i = 0; i < hn; ++i){

 97             scanf("%d %d",&x,&y);

 98             ans += y -  x + 1;

 99             if(i == 0) ans2 += y - x + 3;

100             else{

101                 int mlen = min(y-x,py-px);

102                 if(px == x){

103                     if(mlen == y - x) ans2 += mb(y-x) + 2;

104                     else ans2 += mb(mlen) + 2 + y - x - mlen;

105                 }else{

106                     int com = query(px,x);

107                     if(com >= mlen){

108                         com = mlen;

109                         if(com == y - x) ans2 += mb(com) + 2;

110                         else ans2 += mb(com) + y - x - com + 2;

111                     }else if(com) ans2 += mb(com) + 2 + y - x - com;

112                     else ans2 += y - x + 3;

113                 }

114             }

115             px = x;

116             py = y;

117         }

118         cout<<ans<<" "<<ans2<<endl;

119     }

120     return 0;

121 }

View Code