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POJ 1226 后缀数组 或 KMP 或 暴力

2017-09-13 10:00 441 查看
简略题意:出现或反转后出现在每个字符串中的最长字符串。

先将每个串和自己的反转串连接起来,随后将这若干个串连接起来。二分答案,判定就分组看是否有一组后缀在所有原串或者原串的反转串中出现。

#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 1e5+1100;

int n, q;
int str
;
char str2
;
int belong
;
int f = 0;
int minv;

namespace SA {
int sa
, rank
, height
, s[N<<1], t[N<<1], p
, cnt
, cur
;
#define pushS(x) sa[cur[s[x]]--] = x
#define pushL(x) sa[cur[s[x]]++] = x
#define inducedSort(v) fill_n(sa, n, -1); fill_n(cnt, m, 0);                  \
for (int i = 0; i < n; i++) cnt[s[i]]++;                                  \
for (int i = 1; i < m; i++) cnt[i] += cnt[i-1];                           \
for (int i = 0; i < m; i++) cur[i] = cnt[i]-1;                            \
for (int i = n1-1; ~i; i--) pushS(v[i]);                                  \
for (int i = 1; i < m; i++) cur[i] = cnt[i-1];                            \
for (int i = 0; i < n; i++) if (sa[i] > 0 &&  t[sa[i]-1]) pushL(sa[i]-1); \
for (int i = 0; i < m; i++) cur[i] = cnt[i]-1;                            \
for (int i = n-1;  ~i; i--) if (sa[i] > 0 && !t[sa[i]-1]) pushS(sa[i]-1)
void sais(int n, int m, int *s, int *t, int *p) {
int n1 = t[n-1] = 0, ch = rank[0] = -1, *s1 = s+n;
for (int i = n-2; ~i; i--) t[i] = s[i] == s[i+1] ? t[i+1] : s[i] > s[i+1];
for (int i = 1; i < n; i++) rank[i] = t[i-1] && !t[i] ? (p[n1] = i, n1++) : -1;
inducedSort(p);
for (int i = 0, x, y; i < n; i++) if (~(x = rank[sa[i]])) {
if (ch < 1 || p[x+1] - p[x] != p[y+1] - p[y]) ch++;
else for (int j = p[x], k = p[y]; j <= p[x+1]; j++, k++)
if ((s[j]<<1|t[j]) != (s[k]<<1|t[k])) {ch++; break;}
s1[y = x] = ch;
}
if (ch+1 < n1) sais(n1, ch+1, s1, t+n, p+n1);
else for (int i = 0; i < n1; i++) sa[s1[i]] = i;
for (int i = 0; i < n1; i++) s1[i] = p[sa[i]];
inducedSort(s1);
}
template<typename T>
int mapCharToInt(int n, const T *str) {
int m = *max_element(str, str+n);
fill_n(rank, m+1, 0);
for (int i = 0; i < n; i++) rank[str[i]] = 1;
for (int i = 0; i < m; i++) rank[i+1] += rank[i];
for (int i = 0; i < n; i++) s[i] = rank[str[i]] - 1;
return rank[m];
}
template<typename T>
void suffixArray(int n, const T *str) {
int m = mapCharToInt(++n, str);
sais(n, m, s, t, p);
for (int i = 0; i < n; i++) rank[sa[i]] = i;
for (int i = 0, h = height[0] = 0; i < n-1; i++) {
int j = sa[rank[i]-1];
while (i+h < n && j+h < n && s[i+h] == s[j+h]) h++;
if (height[rank[i]] = h) h--;
}
}
template <typename T>
void init(T *str){
str
= 0;
suffixArray(n, str);
}
int vis[110], count = 0;
void checkinit() {
memset(vis, 0, sizeof vis);
count = 0;
}
bool check(int x) {
checkinit();
for(int i =  1; i <= n; i++) {
if(height[i] >= x) {
int bl = belong[sa[i]];
if(!vis[bl]) vis[bl] = 1, count ++;
if(count == q)
return 1;
} else {
checkinit();
int bl = belong[sa[i]];
if(!vis[bl]) vis[bl] = 1, count ++;
}
}
return 0;
}
void solve() {
int l = 0, r = minv + 1, m;
while(l < r) {
m = (l + r) >> 1;
if(check(m)) l = m + 1;
else r = m;
}
printf("%d\n", l - 1);
}
};

int t;

int main() {
scanf("%d", &t);
while(t--) {
scanf("%d", &q);
int s = 0;
int d = 128;
minv = 0x3f3f3f3f;
for(int i = 1; i <= q; i++) {
scanf("%s", str2);
int len = strlen(str2);
for(int j = 0; j < len; j++) str[s+j] = str2[j], belong[s+j] = i;
s += len;
minv = min(minv, len);
str[s] = d;
d++;
s++;
for(int j = len - 1; j >= 0; j--) str[s+len-1-j] = str2[j], belong[s+j] = i;
s += len;
minv = min(minv, len);
str[s] = d;
d++;
s++;
}
n = s;
SA::init(str);
SA::solve();
}
return 0;
}
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