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SPOJ220---Relevant Phrases of Annihilation(后缀数组+二分,对后缀分组)

2016-09-05 19:18 369 查看
题目

题意:给定N个串,求每个串至少出现两次的最长子串

思路:二分枚举长度,根据长度len分组,若某组里的个数>=k,则说明存在长度为len的至少重复k次子串。

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

const int MAXN = 110000;

char s[MAXN];
int id[MAXN];
int sa[MAXN], rank[MAXN], height[MAXN], c[MAXN], tmp[MAXN];
int n, m, T;

void makesa(int m) {
memset(c, 0, m * sizeof(int));
for(int i = 0; i < n; ++i) ++c[rank[i] = s[i]];
for(int i = 1; i < m; ++i) c[i] += c[i - 1];
for(int i = 0; i < n; ++i) sa[--c[rank[i]]] = i;
for(int k = 1; k < n; k <<= 1) {
for(int i = 0; i < n; ++i) {
int j = sa[i] - k;
if(j < 0) j += n;
tmp[c[rank[j]]++] = j;
}
int j = c[0] = sa[tmp[0]] = 0;
for(int i = 1; i < n; ++i) {
if(rank[tmp[i]] != rank[tmp[i - 1]] || rank[tmp[i] + k] != rank[tmp[i - 1] + k])
c[++j] = i;
sa[tmp[i]] = j;
}
memcpy(rank, sa, n * sizeof(int));
memcpy(sa, tmp, n * sizeof(int));
}
}

void calheight() {
for(int i = 0, k = 0; i < n; height[rank[i++]] = k) {
k -= (k > 0);
int j = sa[rank[i] - 1];
while(s[i + k] == s[j + k]) ++k;
}
}

int mx[MAXN], mn[MAXN];
int stk[MAXN];

void update_max(int &a, int b) {
if(a == -1 || a < b) a = b;
}

void update_min(int &a, int b) {
if(a == -1 || a > b) a = b;
}

bool check(int L) {
int sum = 0, top = 0;
memset(mx, -1, m * sizeof(int));
memset(mn, -1, m * sizeof(int));
memset(c, 0, m * sizeof(int));
for(int i = 0; i < n; ++i) {
if(height[i] >= L) {
update_max(mx[id[sa[i]]], sa[i]);
update_min(mn[id[sa[i]]], sa[i]);
stk[++top] = id[sa[i]];
if(mx[id[sa[i]]] - mn[id[sa[i]]] >= L) {
if(!c[id[sa[i]]]) ++sum;
c[id[sa[i]]] = true;
if(sum >= m) return true;
}
} else {
sum = 0;
while(top) {
int t = stk[top--];
mx[t] = mn[t] = -1;
c[t] = false;
}
update_max(mx[id[sa[i]]], sa[i]);
update_min(mn[id[sa[i]]], sa[i]);
stk[++top] = id[sa[i]];
}
}
return false;
}

int solve() {
int l = 1, r = 5001;
while(l < r) {
int mid = (l + r) >> 1;
if(check(mid)) l = mid + 1;
else r = mid;
}
return l - 1;
}

int main() {
scanf("%d", &T);
while(T--) {
scanf("%d", &m);
n = 0;
for(int i = 0; i < m; ++i) {
scanf("%s", s + n);
while(s
) id[n++] = i;
s[n++] = i + 1;
}
s[n - 1] = 0;
makesa(128);
calheight();
printf("%d\n", solve());
}
}
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