CF 427D Match & Catch 求最短唯一连续LCS

题目来源：CF 427D Match & Catch

题意：给出2个字符串 求最短的连续的公共字符串 并且该字符串在原串中只出现一次

思路：把2个字符串合并起来求height 后缀数组height的应用

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100010;
char s[maxn];
int sa[maxn];
int t[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
int l1, l2;
void build_sa(int m, int n)
{
	int i, *x = t, *y = t2;
	for(i = 0; i < m; i++)
		c[i] = 0;
	for(i = 0; i < n; i++)
		c[x[i] = s[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(int k = 1; k <= n; k <<= 1)
	{
		int p = 0;
		for(i = n-k; i < n; i++)
			y[p++] = i;
		for(i = 0; i < n; i++)
			if(sa[i] >= k)
				y[p++] = sa[i] - k;
		for(i = 0; i < m; i++)
			c[i] = 0;
		for(i = 0; i < n; i++)
			c[x[y[i]]]++;
		for(i = 0; i < m; i++)
			c[i]+= c[i-1];
		for(i = n-1; i >= 0; i--)
			sa[--c[x[y[i]]]] = y[i];
		swap(x,y);
		p = 1; x[sa[0]] = 0;
		for(i = 1; i < n; i++)
			x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
		if(p >= n)
			break;
		m = p;
	}
}

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

bool ok(int m, int n)
{
	int cnt1 = 0, cnt2 = 0;
	for(int i = 1; i <= n; i++)
	{
		if(height[i] >= m)
		{
			if(sa[i-1] < l1)
				cnt1++;
			if(sa[i-1] > l1)
				cnt2++;
		}
		else
		{
			if(sa[i-1] < l1)
				cnt1++;
			if(sa[i-1] > l1)
				cnt2++;
			if(cnt1 == 1 && cnt2 == 1)
				return true;
			cnt1 = cnt2 = 0;
		}
	}
	return false;
}
int main()
{
	char a[5555], b[5555];
	while(scanf("%s %s", &a, &b) != EOF)
	{
		l1 = strlen(a);
		l2 = strlen(b);
		int n = 0;
		for(int i = 0; i < l1; i++)
			s[n++] = a[i];
		s[n++] = ‘z‘+1;
		for(int i = 0; i < l2; i++)
			s[n++] = b[i];
		s[n] = 0;
		build_sa(128, n+1);
		getHeight(n);
		
		int l = 1, r = 5000;
		int ans = -1;
		int len = min(l1, l2);
		for(int i = 1; i <= len; i++)
			if(ok(i, n))
			{
				ans = i;
				break;
			}
		if(ans <= 0)
			printf("-1\n");
		else
			printf("%d\n", ans);
	}
	return 0;
}

CF 427D Match & Catch 求最短唯一连续LCS,布布扣,bubuko.com

CF 427D Match & Catch 求最短唯一连续LCS

原文：http://blog.csdn.net/u011686226/article/details/24937469