[HDU]3518——Boring counting

Posted 2020-09-21 QYP_2002

zhan.jiang.ou now faced a tough problem,his english teacher quan.hong.chun gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other.
Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).

题目大意：求一个字符串的不重叠子串的个数。

思路：我真是太辣鸡了，其实这个题就是后缀数组性质②的应用，只不过是求出每个长度的答案(性质详见我的后缀数组的基本使用技巧)！

总结：后缀数组英语应用太灵活，多回顾反思！

 1 #include<map>
 2 #include<set>
 3 #include<list>
 4 #include<deque>
 5 #include<cmath>
 6 #include<queue>
 7 #include<stack>
 8 #include<vector>
 9 #include<cstdio>
10 #include<complex>
11 #include<cstring>
12 #include<cstdlib>
13 #include<iostream>
14 #include<algorithm>
15 #define LL long long
16 #define db double
17 #define RG register
18 #define maxx 1001
19 #define Inf 666666
20 using namespace std;int tong[maxx];
21 int sa[maxx],X[maxx],Y[maxx],rnk[maxx],height[maxx],num[maxx];char s[maxx];
22 bool comp(int *r,int a,int b,int len){
23   return r[a]==r[b]&&r[a+len]==r[b+len];
24 }
25 void build_sa(int len){
26   int *x=X,*y=Y,*t,Max=maxx/2;
27   memset(tong,0,sizeof(tong));
28   for(int i=0;i<len;++i)
29     tong[x[i]=num[i]]++;
30   for(int i=1;i<=Max;++i)
31     tong[i]+=tong[i-1];
32   for(int i=len-1;i!=-1;i--)
33     sa[--tong[x[i]]]=i;
34   for(int j=1,p=1,i;p<len;j<<=1,Max=p){
35     for(i=len-1,p=0;i>=len-j;--i)y[p++]=i;
36     for(i=0;i<len;++i)
37       if(sa[i]>=j)y[p++]=sa[i]-j;
38     for(i=0;i<=Max;++i)
39       tong[i]=0;
40     for(i=0;i<len;++i)
41       tong[x[y[i]]]++;
42     for(i=1;i<Max;++i)
43       tong[i]+=tong[i-1];
44     for(i=len-1;i!=-1;i--)
45       sa[--tong[x[y[i]]]]=y[i];
46     for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<len;++i)
47       x[sa[i]]=comp(y,sa[i],sa[i-1],j)?p-1:p++;
48   }
49 }int n;
50 void geth(){
51   int j,k=0;
52   for(int i=1;i<=n;++i)
53     rnk[sa[i]]=i;
54   for(int i=0;i<n;height[rnk[i++]]=k)
55     for(k?k--:0,j=sa[rnk[i]-1];num[i+k]==num[j+k];k++);
56 }
57 void ans(){LL ans=0;
58   for(RG int len=1;len<=(n/2);++len){
59     int Max=-Inf,Min=Inf;
60       for(RG int i=1;i<=n;++i){
61     if(height[i]<len){
62       if(Max-Min>=len)ans++;
63       Max=-Inf,Min=Inf;
64     }
65     Max=max(Max,sa[i]);
66     Min=min(Min,sa[i]);
67     if(i==n&&Max-Min>=len)ans++;
68     }
69   }
70   printf("%lld\\n",ans);
71 }
72 int main(){
73   while(scanf("%s",s)){
74     n=strlen(s);
75     if(n==1&&s[0]==\'#\')break;
76     memset(rnk,0,sizeof(rnk));
77     memset(height,0,sizeof(height));
78     memset(sa,0,sizeof(sa));
79     memset(X,0,sizeof(X));
80     memset(Y,0,sizeof(Y));
81     for(int i=0;i<n;++i)
82       num[i]=s[i]-55;
83     num[n]=0;
84     build_sa(n+1);
85     geth();
86     ans();
87   }
88   return 0;
89 }