SPOJ - DISUBSTR Distinct Substrings

最新推荐文章于 2019-08-31 11:55:10 发布

原创最新推荐文章于 2019-08-31 11:55:10 发布 · 246 阅读

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本文介绍了一种求解字符串中不同子串数量的高效算法。通过使用后缀数组技术，该算法能在O(n log n)的时间复杂度内解决问题。文章提供了一个完整的C++实现示例，并解释了核心步骤。

1.题面

http://www.spoj.com/problems/DISUBSTR/en/

2.题意

问一个字符串中有多少个不同的子串

3.思路

参考《后缀数组罗穗蹇》中的思路，很好写

4.代码

/*****************************************************************
    > File Name: Cpp_Acm.cpp
    > Author: Uncle_Sugar
    > Mail: uncle_sugar@qq.com
    > Created Time: 2016年07月29日 星期五 20时13分08秒
*****************************************************************/
# 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;

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 + 200000 ; 
const int INF = INT_MAX>>1;
typedef long long ll;

const int MAXN = 2*100000+10;

int t1[MAXN],t2[MAXN],c[MAXN];

bool cmp(int *r,int a,int b,int l){
	return r[a]==r[b]&&r[a+l]==r[b+l];
}

void da(char str[],int sa[],int rrank[],int height[],int n,int m){
	n++;
	int i, j, p, *x = t1, *y = t2;
	for (i = 0;i < m;i++) c[i] = 0;
	for (i = 0;i < n;i++) c[x[i] = str[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 (j = 1;j <= n; j<<= 1){

		p = 0;
		for (i = n-j; i < n; i++)	y[p++] = i;
		for (i = 0; i < n; i++)	if (sa[i] >= j) y[p++] = sa[i] - j; 

		for (i = 0; i < m; i++) c[i] = 0;
		for (i = 0; i < n; i++) c[x[y[i]]]++;
		for (i = 1; 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]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
		}
		if ( p>=n ) break;
		m = p;
	}
	int k = 0;
	n--;
	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 rrank[MAXN],height[MAXN];
int sa[MAXN];

char str[size];

int main()
{
	std::ios::sync_with_stdio(false);cin.tie(0);
	int i,j;
	int T;
	cin >> T;
	while (T--){
		cin >> str;
		int len = strlen(str); 
		//# void da(char str[],int sa[],int rrank[],int height[],int n,int m){
		da(str,sa,rrank,height,len,128);

		//# for (i=1;i<=len;i++){
			//# cout << "*" << i << "* ";
			//# PrintArray(str+sa[i],str+len);
		//# }
		//# cout << endl;

		ll ans = len - sa[1];
		//# cout << "ans = " << ans << endl;
		for (i=2;i<=len;i++){
			ans += len - sa[i] - height[i];
		}	
		cout << ans << endl;
	}
	return 0;
}