HDU 5769 Substring（后缀数组）

题目链接：HDU 5769

题面：

Substring

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 539    Accepted Submission(s): 226

Problem Description

?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings.
But ?? thinks that is too easy, he wants to make this problem more interesting.
?? likes a character X very much, so he wants to know the number of distinct substrings which contains at least one X.
However, ?? is unable to solve it, please help him.

 

Input

The first line of the input gives the number of test cases T;T test cases follow.
Each test case is consist of 2 lines:
First line is a character X, and second line is a string S.
X is a lowercase letter, and S contains lowercase letters(‘a’-‘z’) only.

T<=30
1<=|S|<=10^5
The sum of |S| in all the test cases is no more than 700,000.

 

Output

For each test case, output one line containing “Case #x: y”(without quotes), where x is the test case number(starting from 1) and y is the answer you get for that case.

 

Sample Input

2 a abc b bbb

 

Sample Output

Case #1: 3 Case #2: 3
Hint
In first case, all distinct substrings containing at least one a: a, ab, abc. In second case, all distinct substrings containing at least one b: b, bb, bbb.
 

 

Author

FZU

 

Source

2016 Multi-University Training Contest 4

 

题意：

    给定一个字符串，问该字符串中包含某一字符的不重复子串的数量。

解题：

   后缀数组的大致原理是懂的，自己写是写不出来的，勉强算是会用吧。这题可以先看一下，如何求某字符串的不重复子串的数量。

   ans=sigma(length-sa[i]-height[i])，如何理解呢，sa[i]表示的是字典序排名为i的后缀，它的起始位置，length-sa[i]，即为排名i的后缀的长度，height[i]，是排名为i的串和它之前那个串的公共前缀长度，故length-sa[i]-height[i]，（减去和字典序前一个的公共前缀）即为不重复子串数量。

   而针对这题，需要包含特殊字符，故预先找到各个后缀中，离该后缀左侧最近的特殊字符的位置，综合该后缀不重复子串长度，取得最值。

代码：

#include <iostream>#include <string>#include <cstring>#include <cstdio>#define rep(i,n) for(int i = 0;i < n; i++)#define sz 100005#define LL long longusing namespace std;char ori[sz];int rk[sz],sa[sz],height[sz],w[sz],wa[sz],res[sz],pos[sz];int min(int a,int b){return a<b?a:b;}int max(int a,int b){return a>b?a:b;}void getSa (int len,int up) {int *k = rk,*id = height,*r = res, *cnt = wa;rep(i,up) cnt[i] = 0;rep(i,len) cnt[k[i] = w[i]]++;rep(i,up) cnt[i+1] += cnt[i];for(int i = len - 1; i >= 0; i--) sa[--cnt[k[i]]] = i;int d = 1,p = 0;while(p < len){for(int i = len - d; i < len; i++) id[p++] = i;rep(i,len)if(sa[i] >= d) id[p++] = sa[i] - d;rep(i,len) r[i] = k[id[i]];rep(i,up) cnt[i] = 0;rep(i,len) cnt[r[i]]++;rep(i,up) cnt[i+1] += cnt[i];for(int i = len - 1; i >= 0; i--)sa[--cnt[r[i]]] = id[i];swap(k,r);p = 0;k[sa[0]] = p++;rep(i,len-1) {if(sa[i]+d < len && sa[i+1]+d <len &&r[sa[i]] == r[sa[i+1]]&& r[sa[i]+d] == r[sa[i+1]+d])k[sa[i+1]] = p - 1;else k[sa[i+1]] = p++;}if(p >= len) return ;d *= 2,up = p, p = 0;}}void getHeight(int len) {rep(i,len) rk[sa[i]] = i;height[0] =  0;for(int i = 0,p = 0; i < len - 1; i++) {int j = sa[rk[i]-1];while(i+p < len&& j+p < len&& w[i+p] == w[j+p]) p++;height[rk[i]] = p;p = max(0,p - 1);}}int getSuffix(char s[]) {int len = strlen(s),up = 0;for(int i = 0; i < len; i++) {w[i] = s[i];up = max(up,w[i]);}w[len++] = 0;getSa(len,up+1);getHeight(len);return len;}int main(){    int t,l;char c;LL ans;scanf("%d",&t);for(int ix=1;ix<=t;ix++){   ans=0;       scanf(" %c",&c);   scanf("%s",ori);       getSuffix(ori);   l=strlen(ori);   int j;   for(pos[l]=l,j=l-1;j>=0;j--)   if(ori[j]==c)   pos[j]=j;       else   pos[j]=pos[j+1];   for(int i=1;i<=l;i++)          ans=ans+min(l-sa[i]-height[i],l-pos[sa[i]]);   printf("Case #%d: %lld\n",ix,ans);}return 0;}