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[HDU 5558] Alice's Classified Message (后缀数组+LCP+单调栈)

2016-10-12 23:39 399 查看

链接

HDU 5558

题意

题意较复杂,可看原题。

基本上等同于对一个字符串的若干个位置i求其前方位置j(0 <= j < i)的后缀suf(j)和suf(i)的LCP,多个答案输出最小j。

题解

裸的后缀数组+LCP问题,我加了一个单调栈优化常数。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
#define maxn (1000010)
bool cmp(int *r, int a, int b, int l)
{ return r[a] == r[b] && r[a + l] == r[b + l]; }
int ta[maxn], tb[maxn], bk[maxn];
void da(int *r, int *sa, int n, int m)
{
int i, j, p, *x = ta, *y = tb, *t;
for(i = 0; i < m; i++) bk[i] = 0;
for(i = 0; i < n; i++) bk[x[i] = r[i]]++;
for(i = 1; i < m; i++) bk[i] += bk[i-1];
for(i = 0; i < n; i++) sa[--bk[x[i]]] = i;
for(j = 1, p = 1; p < n; j *= 2, m = p)
{
for(p = 0, i = n - j; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;
for(i = 0; i < m; i++) bk[i] = 0;
for(i = 0; i < n; i++) bk[x[i]]++;
for(i = 1; i < m; i++) bk[i] += bk[i-1];
for(i = n-1; i >= 0; i--) sa[--bk[x[y[i]]]] = y[i];
for(t = x, x = y, y = t, x[sa[0]] = 0, p = 1, i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
}
}
#define rank rrank
int rank[maxn], sa[maxn], height[maxn];
void calheight(int *r, int n)
{
for(int i = 0; i < n; i++) rank[sa[i]] = i;
for(int k = 0, i = 0; i < n; i++)
{
k ? k-- : 0;
if(rank[i] > 0)
while(r[i + k] == r[sa[rank[i] - 1] + k])
k++;
height[rank[i]] = k;
}
}
#define log2 Log2
int log2[maxn], st[maxn][32];
void st_prepare(int arr[], int n)
{
log2[1] = 0;
for(int i = 2; i <= n; i++)
{
log2[i] = log2[i-1];
if((1 << log2[i] + 1) == i)
log2[i]++;
}
for(int i = n-1; i >= 0; i--)
{
st[i][0] = arr[i];
for(int j = 1; i + (1 << j) - 1 < n; j++)
{
st[i][j] = min(st[i][j-1], st[i + (1 << j-1)][j-1]);
}
}
}
int st_query(int l, int r)
{
int k = log2[r - l + 1];
return min(st[l][k], st[r - (1 << k) + 1][k]);
}
char in[maxn];
int r[maxn], n, left_next[maxn], right_next[maxn], _top;
pair<int, int> _stack[maxn];
inline void _push_stack(int x[], int w, int p)
{
while(_top > 0)
{
if(w < _stack[_top].first)
_top--;
else
break;
}
if(!_top) x[p] = -1;
else x[p] = _stack[_top].second;
_stack[++_top] = make_pair(w, p);
}
int T[maxn], K[maxn];
int main()
{
int t, kase = 0;
cin >> t;
while(t--)
{
scanf("%s", in);
for(n = 0; in
; n++)
r
= in
- 'a' + 1;
r[n++] = 0;
da(r, sa, n, 27);
calheight(r, n);

st_prepare(height, n);
_top = 0;
for(int i = 1; i < n; i++)
_push_stack(left_next, sa[i], i);
_top = 0;
for(int i = n-1; i > 0; i--)
_push_stack(right_next, sa[i], i);
printf("Case #%d:\n", ++kase);
printf("%d %d\n", -1, int(in[0]));
int i = 1, j, K, T, rk;
while(i < n-1)
{
K = 0, T = -1;
rk = rank[i];
j = rk;
while(left_next[j] != -1)
{
j = left_next[j];
int lcp = st_query(j+1, rk);
if(lcp < K) break;
if(lcp > K || lcp == K && sa[j] < T)
K = lcp, T = sa[j];
}
j = rk;
while(right_next[j] != -1)
{
j = right_next[j];
int lcp = st_query(rk+1, j);
if(lcp < K) break;
if(lcp > K || lcp == K && sa[j] < T)
K = lcp, T = sa[j];
}

printf("%d %d\n", K ? K : -1, K ? T : int(in[i]));
i += (K ? K : 1);
}
}
return 0;
}
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