Front compression
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1490 Accepted Submission(s): 553
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
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这题只要解决前后两个串求LCP就行,但是要和这两个串的长度比较一下
/*************************************************************************
> File Name: hdu4691.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年04月18日 星期六 12时53分45秒
************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
class SuffixArray
{
public:
static const int N = 112000;
int init[N];
int X[N];
int Y[N];
int Rank[N];
int sa[N];
int height[N];
int buc[N];
int LOG[N];
int dp[N][20];
int size;
void clear()
{
size = 0;
}
void insert(int n)
{
init[size++] = n;
}
bool cmp(int *r, int a, int b, int l)
{
return (r[a] == r[b] && r[a + l] == r[b + l]);
}
void getsa(int m = 256) //m一般为最大值+1
{
init[size] = 0;
int l, p, *x = X, *y = Y, n = size + 1;
for (int i = 0; i < m; ++i)
{
buc[i] = 0;
}
for (int i = 0; i < n; ++i)
{
++buc[x[i] = init[i]];
}
for (int i = 1; i < m; ++i)
{
buc[i] += buc[i - 1];
}
for (int i = n - 1; i >= 0; --i)
{
sa[--buc[x[i]]] = i;
}
for (l = 1, p = 1; l <= n && p < n; m = p, l *= 2)
{
p = 0;
for (int i = n - l; i < n; ++i)
{
y[p++] = i;
}
for (int i = 0; i < n; ++i)
{
if (sa[i] >= l)
{
y[p++] = sa[i] - l;
}
}
for (int i = 0; i < m; ++i)
{
buc[i] = 0;
}
for (int i = 0; i < n; ++i)
{
++buc[x[y[i]]];
}
for (int i = 1; i < m; ++i)
{
buc[i] += buc[i - 1];
}
for (int i = n - 1; i >= 0; --i)
{
sa[--buc[x[y[i]]]] = y[i];
}
int i;
for (swap(x, y), x[sa[0]] = 0, p = 1, i = 1; i < n; ++i)
{
x[sa[i]] = cmp(y, sa[i - 1], sa[i], l) ? p - 1 : p++;
}
}
}
void getheight()
{
int h = 0, n = size;
for (int i = 0; i <= n; ++i)
{
Rank[sa[i]] = i;
}
height[0] = 0;
for (int i = 0; i < n; ++i)
{
if (h > 0)
{
--h;
}
int j =sa[Rank[i] - 1];
for (; i + h < n && j + h < n && init[i + h] == init[j + h]; ++h);
height[Rank[i] - 1] = h;
}
}
//预处理每一个数字的对数，用于rmq，常数优化
void initLOG()
{
LOG[0] = -1;
for (int i = 1; i < N; ++i)
{
LOG[i] = (i & (i - 1)) ? LOG[i - 1] : LOG[i - 1] + 1;
}
}
void initRMQ()
{
initLOG();
int n = size;
int limit;
for (int i = 0; i < n; ++i)
{
dp[i][0] = height[i];
}
for (int j = 1; j <= LOG[n]; ++j)
{
limit = (n - (1 << j));
for (int i = 0; i <= limit; ++i)
{
dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
}
}
int LCP(int a, int b)
{
int t;
a = Rank[a];
b = Rank[b];
if (a > b)
{
swap(a, b);
}
--b;
t = LOG[b - a + 1];
return min(dp[a][t], dp[b - (1 << t) + 1][t]);
}
}SA;
char str[100100];
int main()
{
while (~scanf("%s", str))
{
int n;
scanf("%d", &n);
SA.clear();
int len = strlen(str);
for (int i = 0; i < len; ++i)
{
SA.insert(str[i] - ‘a‘ + 1);
}
SA.getsa(30);
SA.getheight();
SA.initLOG();
SA.initRMQ();
LL ans1 = 0, ans2 = 0;
int lastl, lastr;
for (int i = 1; i <= n; ++i)
{
int l, r;
scanf("%d%d", &l, &r);
ans1 += (r - l + 1);
if (i == 1)
{
ans2 += (r - l + 1) + 2;
lastl = l;
lastr = r - 1;
continue;
}
--r;
int lcp = -1;
if (l != lastl)
{
lcp = SA.LCP(l, lastl);
}
int len1 = lastr - lastl + 1;
int len2 = r - l + 1;
lastr = r;
lastl = l;
if (lcp == -1)
{
lcp = min(len1, len2);
}
if (lcp > min(len1, len2))
{
lcp = min(len1, len2);
}
++ans2;
ans2 += len2 - lcp + 1;
if (lcp == 0)
{
++ans2;
}
else
{
while (lcp)
{
++ans2;
lcp /= 10;
}
}
}
printf("%lld %lld\n", ans1, ans2);
}
return 0;
}