POJ - 1743 Musical Theme

题意:

就是在一串数字中,找到连续的一串数字,出现两次的最大长度是多少,大于5输出5,小于5输出0。当然还有变调,就是可能同时大一或者小一。

分析:以后一个数和前一个数字的差成立数据,即可避免变调的情况。然后二分答案,后缀数组,求最长公共前缀height数组即可。

坑点:

在后面补一个0。

我用的是刘汝佳的板子,一般情况在后面补一个特别小的数字(比数组里内容都小的数字就可以了)

可是这次以差建立数组,会有0的情况,就改了改板子,(见第27,28行)

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
#include<string>
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 3e4 + 5;
const int maxs = 30;
int s[maxn];
int sa[maxn], t[maxn], t2[maxn], c[maxn], n , m;
int d[maxn][maxs];
int height[maxn], rank[maxn];
void build_sa(int m)
{
	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++)     
		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 i, j, k = 0;
	for (i = 0; i < n; i++)	rank[sa[i]] = i;
	for (i = 0; i < n; i++){
		if (k)	k--;
		 j = sa[rank[i] - 1];
		while (i+k<n&&j+k<n&&s[i + k] == s[j + k])	k++;
		height[rank[i]] = k;
	}
}
void RMQ_init()
{
    for(int i = 0 ; i<n; i++)       d[i][0] = sa[i];
    for(int j = 1 ; (1<<j) - 1 <=n ; j++){
        for(int i = 0 ; i + (1<<j) - 1 < n ; i++){
            d[i][j] = min(d[i][j-1] , d[i + (1<< (j - 1))][ j -1 ]);
        }
    }
}
int RMQ(int L , int R)
{
    int k = 0;
    while((1<<(k + 1)) <= R - L + 1)    k++;
    return min(d[L][k] , d[R - (1 << k) + 1][k]);
}
int q[maxn];
bool check(int mid){
	int minsa,maxsa;
	maxsa=minsa=sa[0];
	for (int i=1;i<n;i++){
		if (height[i]<mid){
			maxsa=minsa=sa[i];
		}else{
			maxsa=max(maxsa,sa[i]);
			minsa=min(minsa,sa[i]);
			if (maxsa-minsa>mid){
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	while(scanf("%d",&n)!=EOF&&n){
		for (int i=0;i<n;i++) scanf("%d",&q[i]);
		if (n<=1){
			printf("0\n");
			continue;
		}
		for (int i=0;i<n-1;i++){
			s[i]=q[i+1]-q[i];
			if (s[i]<0) s[i]+=100;
		}
		s[n-1]=0;
		build_sa(200);
	//	for (int i=0;i<n;i++) cout<<sa[i]<<" ";
		for (int i=0;i<n;i++) sa[i]=sa[i+1];
		n--;
		getHeight();
	//	cout<<endl;
	//	for (int i=1;i<n;i++) cout<<i<<" "<<height[i]<<" "<<sa[i]<<endl;
		int l=0,r=n;
		int mid;
		int zans;
		while(l<=r){
			mid=(l+r)/2;
			int ans=check(mid);
		//	cout<<l<<" "<<r<<" "<<mid<<" "<<ans<<endl; 
			if (ans) {
				zans=mid;
				l=mid+1;
			}
			else{
				r=mid-1;
			}
		}
		if (zans>=4) zans+=1;
		else
		zans=0;
		printf("%d\n",zans);
	} 
	return 0;
}

  

 

posted @ 2017-10-08 17:08  晓风微微  阅读(184)  评论(0编辑  收藏  举报