Front compression
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1339 Accepted Submission(s): 496
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?
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
Author
Zejun Wu (watashi)
Source
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解题思路：后缀数组水题。试了两种模版。还是基数排序的快啊。。
板子1：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll long long
#define maxn 100010
using namespace std;
char s[maxn];
int n,k,q;
int rank[maxn],sa[maxn],tmp[maxn],lcp[maxn];//lcp:0-n-1
bool cmp(int x,int y){
if(rank[x]!=rank[y]) return rank[x]<rank[y];
int sx=x+k<=n ? rank[x+k]:-1;
int sy=y+k<=n ? rank[y+k]:-1;
return sx<sy;
}
void build_sa(){
n=strlen(s);
for(int i=0;i<=n;i++){
sa[i]=i;
rank[i]=i<n ? s[i]:-1;
}
for(k=1;k<=n;k<<=1){
sort(sa,sa+n+1,cmp);
tmp[sa[0]]=0;
for(int i=1;i<=n;i++){
tmp[sa[i]]=tmp[sa[i-1]]+(cmp(sa[i-1],sa[i]) ? 1:0);
}
for(int i=0;i<=n;i++) rank[i]=tmp[i];
}
}
void build_lcp(){
n=strlen(s);
//for(int i=0;i<=n;i++) rank[sa[i]]=i;
int h=0;
lcp[0]=0;
for(int i=0;i<n;i++){
int j=sa[rank[i]-1];
if(h>0) h--;
for(;j+h<n&&i+h<n;h++){
if(s[j+h]!=s[i+h]) break;
}
lcp[rank[i]-1]=h;
}
}
int dp[20][maxn],mm[maxn];
void init_RMQ(int n){
mm[0]=-1;
for(int i=1;i<=n;i++){//长度1-n
mm[i]=(i&(i-1)) ? mm[i-1]:mm[i-1]+1;
}
for(int i=0;i<n;i++) dp[0][i]=lcp[i];
for(int i=1;i<=mm[n];i++){
for(int j=0;j+(1<<i)-1<n;j++){
dp[i][j]=min(dp[i-1][j],dp[i-1][j+(1<<i>>1)]);
}
}
}
int RMQ(int x,int y){//[x,y-1]
if(x==y) return n-x;
x=rank[x],y=rank[y];
if(x>y) swap(x,y);
y--;
int l=mm[y-x+1];
return min(dp[l][x],dp[l][y-(1<<l)+1]);
}
void read(){
scanf("%d",&q);
ll sum1=0,sum2=0;
int pl=-1,pr=-1,l,r;
for(int i=0;i<q;i++){
scanf("%d%d",&l,&r);
sum1+=(r-l+1);
if(pl==-1){
sum2+=r-l+1;
}else{
int LCP=RMQ(pl,l);
int ans=min(LCP,min(r-l,pr-pl));
sum2+=(r-l-ans);
if(ans==0) sum2+=1;
else sum2+=(int)log10(ans*1.0)+1;
}
pl=l,pr=r;
}
printf("%I64d %I64d\n",sum1,sum2+2*q);
}
int main(){
while(~scanf("%s",s)){
build_sa();
build_lcp();
init_RMQ(n);
read();
}
return 0;
}
板子2：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll long long
#define maxn 100010
using namespace std;
char s[maxn];
int c[maxn],wa[maxn],wb[maxn],r[maxn];//求SA数组须要的中间变量，不须要赋值
//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,
//除s[n-1]外的全部s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
int n,sa[maxn],lcp[maxn],rank[maxn];
bool cmp(int *r,int a,int b,int l){
return r[a]==r[b]&&r[a+l]==r[b+l];
}
void build_sa(int n,int m){//数组长度，最大数字
for(int i=0;i<=n;i++) r[i]=i<n ? s[i]:0;
n++;
int i,j,p,*x=wa,*y=wb;
//第一轮基数排序。假设s的最大值非常大，可改为高速排序
for(i=0;i<m;i++) c[i]=0;
for(i=0;i<n;i++) c[x[i]=r[i]]++;
for(i=1;i<m;i++) c[i]+=c[i-1];
for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
for(j=1;j<=n;j<<=1){
p=0;
//直接利用sa数组排序第二keyword
for(i=n-j;i<n;i++) y[p++]=i;//后面的j个数第二keyword为空的最小
for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;
//这样数组y保存的就是依照第二keyword排序的结果
//基数排序第一keyword
for(i=0;i<m;i++) c[i]=0;
for(i=0;i<n;i++) c[x[y[i]]]++;
for(i=1;i<m;i++) c[i]+=c[i-1];
for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
//依据sa和x数组计算新的x数组
swap(x,y);
p=1,x[sa[0]]=0;
for(i=1;i<n;i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
if(p>=n) break;
m=p;
}
}
void build_lcp(int n){
int i,j,k=0;
for(i=0;i<=n;i++) rank[sa[i]]=i;
lcp[0]=0;
for(i=0;i<n;i++){
j=sa[rank[i]-1];
if(k) k--;
while(s[i+k]==s[j+k]) k++;
lcp[rank[i]-1]=k;
}
}
int dp[20][maxn],mm[maxn];
void init_RMQ(int n){
mm[0]=-1;
for(int i=1;i<=n;i++){
mm[i]=(i&(i-1)) ? mm[i-1]:mm[i-1]+1;
}
for(int i=0;i<n;i++) dp[0][i]=lcp[i];
for(int i=1;i<=mm[n];i++){
for(int j=0;j+(1<<i)-1<n;j++){
dp[i][j]=min(dp[i-1][j],dp[i-1][j+(1<<i>>1)]);
}
}
}
int RMQ(int x,int y){
if(x==y) return n-x;
x=rank[x],y=rank[y];
if(x>y) swap(x,y);
y--;
int l=mm[y-x+1];
return min(dp[l][x],dp[l][y-(1<<l)+1]);
}
int q;
void read(){
scanf("%d",&q);
ll sum1=0,sum2=0;
int pl=-1,pr=-1,l,r;
for(int i=0;i<q;i++){
scanf("%d%d",&l,&r);
sum1+=(r-l+1);
if(pl==-1){
sum2+=r-l+1;
}else{
int LCP=RMQ(pl,l);
//cout<<i<<":"<<LCP<<endl;
int ans=min(LCP,min(r-l,pr-pl));
//cout<<i<<":"<<ans<<endl;
sum2+=(r-l-ans);
if(ans==0) sum2+=1;
else sum2+=(int)log10(ans*1.0)+1;
}
pl=l,pr=r;
}
printf("%I64d %I64d\n",sum1,sum2+2*q);
}
int main(){
while(~scanf("%s",s)){
n=strlen(s);
build_sa(n,128);
build_lcp(n);
/*for(int i=0;i<n;i++){
cout<<i<<" "<<sa[i]<<" "<<lcp[i]<<endl;
}*/
init_RMQ(n);
read();
}
return 0;
}