HDU - 3518 Boring counting

1.題面

http://acm.hdu.edu.cn/showproblem.php?pid=3518

2.題意

給你一個字符串,要求你找出一個字符串中出現了至少兩次而且互相不重疊的子串的個數

3.思路

建立height數組之後,枚舉長度,記錄每組中height數組sa[i]的最大值和最小值,只要最大值和最小值之差大於k就可以了

4.代碼

/*****************************************************************
    > File Name: cpp_acm.cpp
    > Author: Uncle_Sugar
    > Mail: [email protected]
    > Created Time: Wed 10 Aug 2016 09:49:47 CST
*****************************************************************/
# include <cstdio>
# include <cstring>
# include <cctype>
# include <cmath>
# include <cstdlib>
# include <climits>
# include <iostream>
# include <iomanip>
# include <set>
# include <map>
# include <vector>
# include <stack>
# include <queue>
# include <algorithm>
using namespace std;

# define rep(i,a,b) for (i=a;i<=b;i++)
# define rrep(i,a,b) for (i=b;i>=a;i--)

struct QuickIO{
	QuickIO(){const int SZ = 1<<20;
		setvbuf(stdin ,new char[SZ],_IOFBF,SZ);
		setvbuf(stdout,new char[SZ],_IOFBF,SZ);
	}				//*From programcaicai*//
}QIO;

template<class T>void PrintArray(T* first,T* last,char delim=' '){
    for (;first!=last;first++) cout << *first << (first+1==last?'\n':delim);
}

const int debug = 1;
const int size  = 10 + 10000000 ; 
const int INF = INT_MAX>>1;
typedef long long ll;

/*
1.see the size of the input data before you select your algorithm 
2.cin&cout is not recommended in ACM/ICPC
3.pay attention to the size you defined, for instance the size of edge is double the size of vertex
*/

const int MAXN = 500000 + 1000;    
#define F(x) ((x)/3+((x)%3==1?0:tb))    
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)    
int wa[MAXN*3],wb[MAXN*3],wv[MAXN*3],wss[MAXN*3];    

int c0(int *r,int a,int b){    
	return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];    
}    

int c12(int k,int *r,int a,int b){    
	if(k == 2)    
		return r[a] < r[b] || ( r[a] == r[b] && c12(1,r,a+1,b+1) );    
	else return r[a] < r[b] || ( r[a] == r[b] && wv[a+1] < wv[b+1] );    
}    

void sort(int *r,int *a,int *b,int n,int m){    
	int i;    
	for(i = 0;i < n;i++)wv[i] = r[a[i]];    
	for(i = 0;i < m;i++)wss[i] = 0;    
	for(i = 0;i < n;i++)wss[wv[i]]++;    
	for(i = 1;i < m;i++)wss[i] += wss[i-1];    
	for(i = n-1;i >= 0;i--)    
		b[--wss[wv[i]]] = a[i];    
}    

void dc3(int *r,int *sa,int n,int m){    
	int i, j, *rn = r + n;    
	int *san = sa + n, ta = 0, tb = (n+1)/3, tbc = 0, p;    
	r[n] = r[n+1] = 0;    
	for(i = 0;i < n;i++)if(i %3 != 0)wa[tbc++] = i;    
	sort(r + 2, wa, wb, tbc, m);    
	sort(r + 1, wb, wa, tbc, m);    
	sort(r, wa, wb, tbc, m);    
	for(p = 1, rn[F(wb[0])] = 0, i = 1;i < tbc;i++)    
		rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p - 1 : p++;    
	if(p < tbc)dc3(rn,san,tbc,p);    
	else for(i = 0;i < tbc;i++)san[rn[i]] = i;    
	for(i = 0;i < tbc;i++) if(san[i] < tb)wb[ta++] = san[i] * 3;    
	if(n % 3 == 1)wb[ta++] = n - 1;    
	sort(r, wb, wa, ta, m);    
	for(i = 0;i < tbc;i++)wv[wb[i] = G(san[i])] = i;    
	for(i = 0, j = 0, p = 0;i < ta && j < tbc;p++)    
		sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];    
	for(;i < ta;p++)sa[p] = wa[i++];    
	for(;j < tbc;p++)sa[p] = wb[j++];    
}    
//str和sa也要三倍    

void da(int str[],int sa[],int rrank[],int height[],int n,int m)    
{    
	for(int i = n;i < n*3;i++)    
		str[i] = 0;    
	dc3(str, sa, n+1, m);    
	int i,j,k = 0;    
	for(i = 0;i <= n;i++)rrank[sa[i]] = i;    
	for(i = 0;i < n; i++)    
	{    
		if(k) k--;    
		j = sa[rrank[i]-1];    
		while(str[i+k] == str[j+k]) k++;    
		height[rrank[i]] = k;    
	}    
}    

int sa[MAXN],rrank[MAXN],height[MAXN];    

int n,k;        

int str[MAXN],len;   
char s[MAXN];  

void Debug(int n){  
	for (int i=1;i<=n;i++){  
		cout << "*" << i << "*\t" << height[i] << "\t";
		PrintArray(str+sa[i],str+n,'\t');  
	}  
}  

int main()
{
	int i,j;
	while (scanf("%s",s)){
		if (s[0]=='#') break;
		for (n=0;s[n]!='\0';n++) str[n] = s[n];
		//# void da(int str[],int sa[],int rrank[],int height[],int n,int m)    
		da(str,sa,rrank,height,n,250);

		int cnt = 0;
		for (k=1;k<=n;k++){
			int mi = INF, ma = -INF;
			for (i=1;i<=n;i++){
				if (height[i] >= k){
					ma = max(sa[i],ma);
					mi = min(sa[i],mi);
				}else {
					if (ma - mi >= k)	
						cnt++;
					mi = ma = sa[i];
				}
			}
			if (ma - mi >= k)	
				cnt++;
		}
		printf("%d\n",cnt);
		
	
	}
	return 0;
}