Light OJ 1314 Names for Babies

Posted 2020-08-02 GFY

http://www.lightoj.com/volume_showproblem.php?problem=1314

题意:给定一个串和p,q，求长度在p到q之间的子串有几种

思路:后缀数组，对于每个位置的贡献是min(n-sa[i],q)，然后要减去重复和没有的部分，就是max(height[i],p-1)，当然，还要对0取个max

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 char s[200005];
 7 int num[200005],sa[200005],rank[200005],h[200005],p,q,n;
 8 int ws[200005],wa[200005],wb[200005],wv[200005];
 9 int read(){
10     int t=0,f=1;char ch=getchar();
11     while (ch<‘0‘||ch>‘9‘){if (ch==‘-‘)f=-1;ch=getchar();}
12     while (‘0‘<=ch&&ch<=‘9‘){t=t*10+ch-‘0‘;ch=getchar();}
13     return t*f;
14 }
15 void solve(){
16     int ans=0;
17     for (int i=1;i<=n;i++)
18      ans+=std::max(0,std::min(q,n-sa[i])-std::max(p-1,h[i]));
19     printf("%d\n",ans); 
20 }
21 bool cmp(int *r,int a,int b,int l){
22     return r[a]==r[b]&&r[a+l]==r[b+l];
23 }
24 void da(int *r,int *sa,int n,int m){
25     int *x=wa,*y=wb,*t,i,j,p;
26     for (i=0;i<n;i++) x[i]=r[i];
27     for (i=0;i<m;i++) ws[i]=0;
28     for (i=0;i<n;i++) ws[x[i]]++;
29     for (i=1;i<m;i++) ws[i]+=ws[i-1];
30     for (i=n-1;i>=0;i--) sa[--ws[x[i]]]=i;
31     for (j=1,p=1;p<n;m=p,j*=2){
32         for (p=0,i=n-j;i<n;i++) y[p++]=i;
33         for (i=0;i<n;i++) if (sa[i]>=j) y[p++]=sa[i]-j;
34         for (i=0;i<n;i++) wv[i]=x[y[i]];
35         for (i=0;i<m;i++) ws[i]=0;
36         for (i=0;i<n;i++) ws[wv[i]]++;
37         for (i=1;i<m;i++) ws[i]+=ws[i-1];
38         for (i=n-1;i>=0;i--) sa[--ws[wv[i]]]=y[i];
39         for (t=x,x=y,y=t,i=1,x[sa[0]]=0,p=1;i<n;i++)
40          x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
41     }
42 }
43 void cal(int *r,int n){
44     int i,j,k=0;
45     for (i=1;i<=n;i++) rank[sa[i]]=i;
46     for (i=0;i<n;h[rank[i++]]=k)
47      for (k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
48 }
49 int main(){
50     int T=read(),Tcase=0;
51     while (T--){
52         Tcase++;printf("Case %d: ",Tcase);
53         scanf("%s",s);
54         p=read();q=read();
55         n=strlen(s);
56         for (int i=0;i<n;i++) num[i]=s[i];
57         num[n]=0;
58         da(num,sa,n+1,200);
59         cal(num,n);
60         solve();
61     }
62 }