本文主要是介绍SPOJ - DISUBSTR Distinct Substrings,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
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;
}
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