HDU-4691 Front compression 后缀数组

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4691

　　后缀数组模板题，求出Height数组后，对Height做RMQ，然后直接统计就可以了。。。

 //STATUS:C++_AC_828MS_11284KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 char s[N];
 int d[N][];
 int num[N];
 int sa[N],t1[N],t2[N],c[N],rank[N],height[N];
 int n,m;

 void build_sa(int s[],int n,int m)
 {
     int i,k,p,*x=t1,*y=t2;
     //第一轮基数排序
     for(i=;i<m;i++)c[i]=;
     for(i=;i<n;i++)c[x[i]=s[i]]++;
     for(i=;i<m;i++)c[i]+=c[i-];
     for(i=n-;i>=;i--)sa[--c[x[i]]]=i;
     for(k=;k<=n;k<<=){
         p=;
         //直接利用sa数组排序第二关键字
         for(i=n-k;i<n;i++)y[p++]=i;
         for(i=;i<n;i++)if(sa[i]>=k)y[p++]=sa[i]-k;
         //基数排序第一关键字
         for(i=;i<m;i++)c[i]=;
         for(i=;i<n;i++)c[x[y[i]]]++;
         for(i=;i<m;i++)c[i]+=c[i-];
         for(i=n-;i>=;i--)sa[--c[x[y[i]]]]=y[i];
         //根据sa和x数组计算新的x数组
         swap(x,y);
         p=;x[sa[]]=;
         for(i=;i<n;i++)
             x[sa[i]]=y[sa[i-]]==y[sa[i]] && y[sa[i-]+k]==y[sa[i]+k]?p-:p++;
         if(p>=n)break;   //已经排好序，直接退出
         m=p;     //下次基数排序的最大值
     }
 }

 void getHeight(int s[],int n)
 {
     int i,j,k=;
     for(i=;i<=n;i++)rank[sa[i]]=i;
     for(i=;i<n;i++){
         if(k)k--;
         j=sa[rank[i]-];
         while(s[i+k]==s[j+k])k++;
         height[rank[i]]=k;
     }
 }

 void rmq_init(int a[])
 {
     int i,j;
     for(i=;i<=n;i++)d[i][]=a[i];
     for(j=;(<<j)<=n;j++){
         for(i=;i+(<<j)-<=n;i++){
             d[i][j]=Min(d[i][j-],d[i+(<<(j-))][j-]);
         }
     }
 }

 int rmq(int l,int r)
 {
     int k=;
     while((<<(k+))<=r-l+)k++;
     return Min(d[l][k],d[r-(<<k)+][k]);
 }

 int lcp(int a,int b)
 {
     if(a==b)return n-a;      //a和b为同一后缀，直接输出，字串串长度为n
     int ra=rank[a],rb=rank[b];
     if(ra>rb)swap(ra,rb);
     return rmq(ra+,rb);
 }

 int w[N];

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,k,Q,a,b,la,lb;
     LL ans1,ans2,t;
     w[]=;
     for(i=k=;i<N;i=j,k++)
         for(j=i*;i<j && i<N;i++)w[i]=k;
     while(~scanf("%s",s))
     {
         n=strlen(s);
         for(i=;i<n;i++)
             num[i]=s[i]-'a'+;
         num[n]=;m=;
         build_sa(num,n+,m);
         getHeight(num,n);
         rmq_init(height);

         scanf("%d",&Q);
         ans1=(LL)Q,ans2=(LL)*Q;
         scanf("%d%d",&la,&lb);
         ans1+=(LL)lb-la;ans2+=(LL)lb-la+;
         while(--Q){
             scanf("%d%d",&a,&b);
             ans1+=(LL)b-a;
             t=(LL)Min(lcp(la,a),lb-la,b-a);
             ans2+=(LL)b-a-t+w[t];
         //    printf("  %I64d %d %d %d\n",t,lcp(la,a),lb-la,b-a);
             la=a;lb=b;
         }

         printf("%I64d %I64d\n",ans1,ans2);

     }
     return ;
 }