POJ 1743 Musical Theme 后缀数组 不可重叠最长重复子串

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二分长度k 长度大于等于k的分成一组 每组sa最大的和最小的距离大于k 说明可行

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 20010;
int s[maxn];
int sa[maxn];
int t[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
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 n, int k)
{
	int ans1 = sa[1], ans2 = sa[1];
	for(int i = 2; i <= n; i++)
	{
		if(height[i] < k)
		{
			ans1 = ans2 = sa[i];
		}
		else
		{
			ans1 = max(ans1, sa[i]);
			ans2 = min(ans2, sa[i]);
			if(ans1-ans2 > k)
				return true;
		}
	}
	return false;
}
int main()
{
	int n;
	while(scanf("%d", &n) && n)
	{
		for(int i = 0; i < n; i++)
			scanf("%d", &s[i]);
		n--;
		for(int i = 0; i < n; i++)
			s[i] = s[i+1] - s[i] + 90;
		s[n] = 0;
		build_sa(200, n+1);
	
		getHeight(n);
		int l = 4, r = n+1;
		int ans = -1;
		while(l <= r)
		{
			int m = (l+r) >> 1;
			if(ok(n, m))
			{
				l = m+1;
				ans = m;
			}
			else
				r = m-1;
		}
		printf("%d\n", ans+1);
	}
	return 0;
}

POJ 1743 Musical Theme 后缀数组 不可重叠最长重复子串

标签：blog   io   sp   for   on   2014   log   html   amp   

原文地址：http://blog.csdn.net/u011686226/article/details/41946197