HDU 4691 Front compression
Front compression
Time Limit: 5000ms
Memory Limit: 102400KB
This problem will be judged on HDU. Original ID: 469164-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:
![](http://www.bnuoj.com/v3/images/hdu/C463-1006-1.jpg)
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?
![](http://www.bnuoj.com/v3/images/hdu/C463-1006-1.jpg)
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.
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
解题:后缀数组配合RMQ的使用
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }
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