題目鏈接：HDU 6194
string string string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 671 Accepted Submission(s): 196
Problem Description
Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.
Input
The first line contains an integer T (T≤100)
implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.
Output
For each test case, print the number of the important substrings in a line.
Sample Input
2
2
abcabc
3
abcabcabcabc
Sample Output
6
9
Source
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題目分析：詢問不同子串個數那基本就是後綴數組沒跑了，這裏將所有的子串按排名列出來，可以看出，排名連續的子串的公共前綴一定出現了至少k次。之後我們的工作就是判斷這裏有多找個前綴出現次數比k次多，然後分別排名往前和往後看一位，如果還有公共前綴就把這些前綴去掉，height數組保存的就是牌名相鄰的2個串公共前綴，我們是滑動大小爲k的窗口並維護最小值，可以單調棧可以RMQ，不過後綴數組模板自帶RMQ，調整下RMQ查詢即可。
細節問題：
我的模板height[i]保存的是排名i與排名i-1的公共前綴，所以要注意查詢範圍。
rank數組從1開始，height數組也是從1開始。
RMQ模板傳的2個參數是左右邊界在實際字符串的位置，所以對於排名i需要傳sa[i]。
對於K=1的情況不特判會RE（RE-1），實際上k=1相當於窗口大小爲1，那麼公共前綴就是自己的長度，其他相同，不需要添油加醋（WA-2）TAT。
PS：對於第9題，我們隊一早就發現了這個bug，然後有一段時間忙着寫題沒管，然後等卡題了於是我們閒着無聊就二進制從1開始挨個試着交，然後交了200多次竟然過了。。。然後吃瓜羣衆看到我們原來還有這種操作，就開始也暴力提交，，，於是boom！！！真是打得最歡樂的一場網絡賽了，整個比賽節奏都被我們帶歪了。。
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int K;
const int maxn=200010;
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(int str[],int sa[],int ranks[],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++)ranks[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k)k--;
j=sa[ranks[i]-1];
while(str[i+k]==str[j+k])k++;
height[ranks[i]]=k;
}
}
int ranks[maxn],height[maxn];
char str[maxn];
int r[maxn],sa[maxn];
int pre[maxn],vis[maxn];
int RMQ[maxn],mm[maxn];
int dp[maxn][60];
void Rmq_Init(int n)
{
int m=floor(log(n+0.0)/log(2.0));
for(int i=1;i<=n;i++) dp[i][0]=height[i];
for(int i=1;i<=m;i++)
{
for(int j=n;j;j--){
dp[j][i]=dp[j][i-1];
if(j+(1<<(i-1))<=n)
dp[j][i]=min(dp[j][i],dp[j+(1<<(i-1))][i-1]);
}
}
}
int Rmq_Query(int l,int r)
{
int a=ranks[l],b=ranks[r];
if(a>b) swap(a,b);
a++;
int m=floor(log(b-a+1.0)/log(2.0));
return min(dp[a][m],dp[b-(1<<m)+1][m]);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&K);
scanf("%s",str);
int len=strlen(str);
for(int i=0;i<len;i++) r[i]=str[i];
r[len]=0;
da(r,sa,ranks,height,len,128);
Rmq_Init(len);
long long int ans=0;
for(int i=1;i+K-1<=len;i++)
{
int buf2=0,buf3=0,buf1=0;
int a1=sa[i],a2=sa[i+K-1];
if(K!=1) buf1=Rmq_Query(a1, a2);
else buf1=len-sa[i];
if(i!=0) buf2=height[i];
if(i+K-1<len) buf3=height[i+K];
if(buf2>buf1||buf3>buf1) continue;
else
{
ans+=buf1-max(buf3,buf2);
}
}
printf("%lld\n",ans);
}
}
