HDU 5769 Substring 后缀数组

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原文链接：http://www.cnblogs.com/zxhl/p/6002089.html

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#数据结构与算法

本文介绍了一种使用后缀数组解决含特定字符的子串计数问题的方法。通过构建后缀数组和高度数组，结合预处理技巧，有效地统计了所有包含特定字符的不同子串的数量。

Substring

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.

题意：

　　给你一个字符c和一个字符串s

　　问你 s的所有有多少个不同的子串是包含c的

题解：

　　考虑后缀数组

　　我们对S建立出SA,Height

　　在sa中，我们想要知道当前sa[i] 的后缀中是否含有字符c，这个可以预处理呀

　　处理后，假设对于当前sa[i]，的最前一个字符c的位置存在设为pre[sa[i]]，那么我们可以更新答案 n - sa[i]

　　但题目要求得是不同子串，那么我们对于每一位sa[i],利用height数组去重就可以了，自己想想把

#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 2e5+10, M = 2e5+20, mod = 1e9+7, inf = 2e9;

///heght[i] 表示 Suffix(sa[i-1])和Suffix(sa[i]) 的最长公共前缀：
///rk[i] 表示 开头为i的后缀的等级：
///sa[i] 表示 排名为i的后缀 的开头位置：

int *rk,r[N],sa[N],height[N],wa[N],wb[N],wm[N];
bool cmp(int *r,int a,int b,int l) {
    return r[a] == r[b] && r[a+l] == r[b+l];
}
void SA(int *r,int *sa,int n,int m) {
    int *x=wa,*y=wb,*t;
    for(int i=0;i<m;++i)wm[i]=0;
    for(int i=0;i<n;++i)wm[x[i]=r[i]]++;
    for(int i=1;i<m;++i)wm[i]+=wm[i-1];
    for(int i=n-1;i>=0;--i)sa[--wm[x[i]]]=i;
    for(int i=0,j=1,p=0;p<n;j=j*2,m=p){
        for(p=0,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)wm[i]=0;
        for(i=0;i<n;++i)wm[x[y[i]]]++;
        for(i=1;i<m;++i)wm[i]+=wm[i-1];
        for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i];
        for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i) {
            x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;
        }
    }
    rk=x;
}
void Height(int *r,int *sa,int n) {
    for(int i=0,j=0,k=0;i<n;height[rk[i++]]=k)
    for(k?--k:0,j=sa[rk[i]-1];r[i+k] == r[j+k];++k);
}

int pre[N];
LL ans;
char ch[2],s[N];
int main() {
        int T,cas = 1;
        scanf("%d",&T);
        while(T--) {
            scanf("%s%s",ch,s);
            int n = strlen(s);
            for(int i = 0; i < n; ++i) r[i] = s[i] - 'a' + 1;
            r[n] = 0;
            SA(r,sa,n+1,256);
            Height(r,sa,n);
            pre[n] = -1;
            for(int i = n-1; i >= 0; --i) {
                if(ch[0] == s[i]) pre[i] = i;
                else pre[i] = pre[i+1];
            }
            ans = 0;
            for(int i = 1; i <= n; ++i) {
                if(pre[sa[i]] == -1) continue;
                ans = ans + (n - pre[sa[i]]) -  max(0,height[i]  -  pre[sa[i]] + sa[i]);
            }
            printf("Case #%d: %I64d\n",cas++,ans);
        }
        return 0;
}