[HDU]3518 Boring counting 做题笔记
来源:互联网 发布:淘宝文具盒 编辑:程序博客网 时间:2024/05/20 16:02
题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=3518
这题和POJ1743是差不多的类型。
把二分改为穷举长度l,分块(如果lcp>=l则分到一个块),一个块里若存在两个位置差>=l的则ans++
很好奇这题的这句话是摆来吓唬人的??
For each test case output an integer ans,which represent the answer for the test case.You’d better use int64 to avoid unnecessary trouble.
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N=2000;int r[N],wa[N],wb[N],wv[N],wd[N];int sa[N],rk[N],height[N];int ans=0,n;char s[N];bool cmp (int *r,int a,int b,int l) { return r[a]==r[b] && r[a+l]==r[b+l];}void da (int *r,int *sa,int n,int m) { int i,j,p,*x=wa,*y=wb,*t; for (i=0;i<m;i++) wd[i]=0; for (i=0;i<n;i++) wd[x[i]=r[i]]++; for (i=1;i<m;i++) wd[i]+=wd[i-1]; for (i=n-1;i>=0;i--) sa[--wd[x[i]]]=i; for (j=1,p=1;p<n;j<<=1,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<n;i++) wv[i]=x[y[i]]; for (i=0;i<m;i++) wd[i]=0; for (i=0;i<n;i++) wd[wv[i]]++; for (i=1;i<m;i++) wd[i]+=wd[i-1]; for (i=n-1;i>=0;i--) sa[--wd[wv[i]]]=y[i]; for (t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; }}void calheight (int *r,int *sa,int n) { int i,j,k=0; for (i=1;i<=n;i++) rk[sa[i]]=i; for (i=0;i<n;height[rk[i++]]=k) for (k?k--:0,j=sa[rk[i]-1];r[i+k]==r[j+k];k++);}int work (int x) { int i=2,mi,ma,ans=0; while (1) { while (i<=n&&height[i]<x) i++; if (i>n) break; mi=ma=sa[i-1];//注意:height[i]是i-1和i的lcp,所以不能少了i-1 while (i<=n&&height[i]>=x) { mi=min(mi,sa[i]); ma=max(ma,sa[i]); i++; } if (ma-mi>=x) ans++; } return ans;}int main () { while (scanf("%s",s)!=EOF) { if (strcmp(s,"#")==0) return 0; n=strlen(s);ans=0; for (int i=0;i<n;i++) r[i]=s[i]; r[n]=0; da(r,sa,n+1,128); calheight(r,sa,n); for (int i=1;i<=n/2;i++) ans+=work(i); printf("%d\n",ans); }}
0 0
- [HDU]3518 Boring counting 做题笔记
- hdu 3518 Boring counting
- Boring counting HDU 3518
- hdu 3518 Boring counting
- hdu-3518-Boring counting
- HDU 3518 Boring counting
- HDU 3518 Boring counting
- hdu 3518 Boring counting
- HDU 3518 Boring counting
- HDU - 3518 Boring counting
- HDU 3518Boring counting
- HDU 3518 Boring counting(后缀数组入门题)
- HDU 3518 Boring counting(后缀数组)
- HDU 3518 Boring counting - 后缀树
- HDU 3518 Boring counting && 后缀数组
- 【HDU 3518】Boring counting【扩展KMP】
- hdu 3518 Boring counting(后缀数组)
- [后缀数组+枚举] hdu 3518 Boring counting
- Java你可能不知道的事(3)HashMap
- LinearLayout(线性布局)
- 谁将会参加比赛的问题(C实现)
- 磁力链接 结构解析 分享
- 死锁的处理
- [HDU]3518 Boring counting 做题笔记
- iptables详解
- UILabel的minimumScaleFactor
- 仿QQ侧滑删除Item:Swipemenulistview的简单实现
- JavaWeb项目如何在局域网内发布
- javascript中关于冒泡事件的常用方法
- Activity的跳转
- 【HDU】 5538 House Building
- SDUT 3510 快速幂