zoukankan      html  css  js  c++  java
  • HDU 4416 Good Article Good sentence

    Description

    In middle school, teachers used to encourage us to pick up pretty sentences so that we could apply those sentences in our own articles. One of my classmates ZengXiao Xian, wanted to get sentences which are different from that of others, because he thought the distinct pretty sentences might benefit him a lot to get a high score in his article.
    Assume that all of the sentences came from some articles. ZengXiao Xian intended to pick from Article A. The number of his classmates is n. The i-th classmate picked from Article Bi. Now ZengXiao Xian wants to know how many different sentences she could pick from Article A which don't belong to either of her classmates?Article. To simplify the problem, ZengXiao Xian wants to know how many different strings, which is the substring of string A, but is not substring of either of string Bi. Of course, you will help him, won't you?

    solution

    以A建立SAM
    在字符串中一个节点 (p) 可以接受的子串为 (len[p]),但是有一些被父亲节点接受了,所以他实际只可以接受 (len[p]-len[fa[p]]),所以我们拿 (bi...bn) 在上面跑,记录一个在一个节点最多被匹配的长度 (f[i]),每一个节点的贡献就是 (Max(len[i]-Max(f[i],len[fa[i]]),0)),注意 (f[i]) 要向父亲上传

    #include <algorithm>
    #include <iostream>
    #include <cstdlib>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    #define RG register
    #define il inline
    #define iter iterator
    #define Max(a,b) ((a)>(b)?(a):(b))
    #define Min(a,b) ((a)<(b)?(a):(b))
    using namespace std;
    const int N=200005;
    char s[N];int m,n,cnt=1,len[N],ch[N][27],p,cur=1,fa[N];
    void build(int c,int id){
       p=cur;cur=++cnt;len[cur]=id;
       for(;p && !ch[p][c];p=fa[p])ch[p][c]=cur;
       if(!ch[p][c])fa[cur]=1;
       else{
          int q=ch[p][c];
          if(len[p]+1==len[q])fa[cur]=q;
          else{
             int nt=++cnt;len[nt]=len[p]+1;
             memcpy(ch[nt],ch[q],sizeof(ch[q]));
             fa[nt]=fa[q];fa[q]=fa[cur]=nt;
             for(;ch[p][c]==q;p=fa[p])ch[p][c]=nt;
          }
       }
    }
    int kase=0,f[N];
    void matcher(){
       int c,l=strlen(s+1),p=1,x=0;
       for(int i=1;i<=l;i++){
          c=s[i]-'a';
          if(!ch[p][c]){
             while(p && !ch[p][c])p=fa[p];
             if(!p)p=1,x=0;
             else x=len[p]+1,p=ch[p][c];
          }
          else x++,p=ch[p][c];
          if(x>f[p])f[p]=x;
       }
    }
    int c[N],sa[N];
    void Clear(){
       memset(c,0,sizeof(c));cnt=1;cur=1;
       memset(ch,0,sizeof(ch));memset(f,0,sizeof(f));
       memset(fa,0,sizeof(fa));
    }
    void work()
    {
       Clear();
       scanf("%d%s",&m,s+1);
       n=strlen(s+1);
       for(int i=1;i<=n;i++)build(s[i]-'a',i);
       for(int i=1;i<=m;i++){
          scanf("%s",s+1);
          matcher();
       }
       for(int i=1;i<=cnt;i++)c[len[i]]++;
       for(int i=1;i<=n;i++)c[i]+=c[i-1];
       for(int i=cnt;i;i--)sa[c[len[i]]--]=i;
       for(int i=cnt,x;i;i--){
          x=sa[i];
          f[fa[x]]=Max(f[fa[x]],f[x]);
       }
       long long ans=0,tmp;
       for(int i=1;i<=cnt;i++){
          tmp=Max(f[i],len[fa[i]]);
          ans+=Max(len[i]-tmp,0);
       }
       printf("Case %d: %lld
    ",++kase,ans);
    }
    
    int main()
    {
    	int T;cin>>T;
            while(T--)work();
    	return 0;
    }
    
    
  • 相关阅读:
    面向过程(或者叫结构化)分析方法与面向对象分析方法到底区别在哪里?请根据自己的理解简明扼要的回答
    当下大部分互联网创业公司为什么都愿意采用增量模型来做开发?
    0
    计算机网络
    java基础
    java 多线程编程
    java类与对象,用程序解释
    修饰符的探讨
    java学习总结02
    java day1
  • 原文地址:https://www.cnblogs.com/Yuzao/p/7672461.html
Copyright © 2011-2022 走看看