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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 100010;
int rk[maxn],wb[maxn],wv[maxn],wd[maxn],lcp[maxn];
bool cmp(int *r,int i,int j,int k){
    return r[i] == r[j] && r[i+k] == r[j+k];
}
void da(int *r,int *sa,int n,int m){
    int i,k,p,*x = rk,*y = wb;
    for(i = 0; i < m; ++i) wd[i] = 0;
    for(i = 0; i < n; ++i) wd[x[i] = r[i]]++;
    for(i = 1; i < m; ++i) wd[i] += wd[i-1];
    for(i = n-1; i >= 0; --i) sa[--wd[x[i]]] = i;

    for(p = k = 1; p < n; k <<= 1,m = p){
        for(p = 0,i = n-k; i < n; ++i) y[p++] = i;
        for(i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < n; ++i) wv[i] = x[y[i]];

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

        swap(x,y);
        x[sa[0]] = 0;
        for(p = i = 1; i < n; ++i)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],k)?p-1:p++;
    }
}
void calcp(int *r,int *sa,int n){
    for(int i = 1; i <= n; ++i) rk[sa[i]] = i;
    int h = 0;
    for(int i = 0; i < n; ++i){
        if(h > 0) h--;
        for(int j = sa[rk[i]-1]; i+h < n && j+h < n; h++)
            if(r[i+h] != r[j+h]) break;
        lcp[rk[i]] = h;
    }
}
int st[maxn][20];
void init(int n){
    memset(st,0,sizeof(st));
    for(int i = 1; i <= n; ++i) st[i][0] = lcp[i];
    for(int i = 1; 1<<i <= n; ++i){
        for(int j = 1; j+(1<<i) <= n+1; ++j)
            st[j][i] = min(st[j][i-1],st[j+(1<<(i-1))][i-1]);
    }
}
int query(int s,int t){
    s = rk[s];
    t = rk[t];
    if(s > t) swap(s,t);
    s++;
    int k = log2(t - s + 1);
    return min(st[s][k],st[t-(1<<k)+1][k]);
}
int r[maxn],sa[maxn];
char str[maxn];
int mb(int x){
    int ans = 0;
    if(x == 0) return 1;
    while(x){
        x /= 10;
        ++ans;
    }
    return ans;
}
int main() {
    int hn,x,y;
    while(~scanf("%s",str)){
        int len = strlen(str);
        for(int i = 0; str[i]; ++i)
            r[i] = str[i];
        r[len] = 0;
        da(r,sa,len+1,128);
        calcp(r,sa,len);
        init(len);
        LL ans = 0,ans2 = 0;
        int px,py;
        scanf("%d",&hn);
        for(int i = 0; i < hn; ++i){
            scanf("%d %d",&x,&y);
            ans += y -  x + 1;
            if(i == 0) ans2 += y - x + 3;
            else{
                int mlen = min(y-x,py-px);
                if(px == x){
                    if(mlen == y - x) ans2 += mb(y-x) + 2;
                    else ans2 += mb(mlen) + 2 + y - x - mlen;
                }else{
                    int com = query(px,x);
                    if(com >= mlen){
                        com = mlen;
                        if(com == y - x) ans2 += mb(com) + 2;
                        else ans2 += mb(com) + y - x - com + 2;
                    }else if(com) ans2 += mb(com) + 2 + y - x - com;
                    else ans2 += y - x + 3;
                }
            }
            px = x;
            py = y;
        }
        cout<<ans<<" "<<ans2<<endl;
    }
    return 0;
}