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。。。

  1 //STATUS:C++_AC_828MS_11284KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //#pragma comment(linker,"/STACK:102400000,102400000")

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef __int64 LL;

 34 typedef unsigned __int64 ULL;

 35 //const

 36 const int N=100010;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=1000000007,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-8;

 41 const double OO=1e15;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 char s[N];

 59 int d[N][20];

 60 int num[N];

 61 int sa[N],t1[N],t2[N],c[N],rank[N],height[N];

 62 int n,m;

 63 

 64 void build_sa(int s[],int n,int m)

 65 {

 66     int i,k,p,*x=t1,*y=t2;

 67     //第一轮基数排序

 68     for(i=0;i<m;i++)c[i]=0;

 69     for(i=0;i<n;i++)c[x[i]=s[i]]++;

 70     for(i=1;i<m;i++)c[i]+=c[i-1];

 71     for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;

 72     for(k=1;k<=n;k<<=1){

 73         p=0;

 74         //直接利用sa数组排序第二关键字

 75         for(i=n-k;i<n;i++)y[p++]=i;

 76         for(i=0;i<n;i++)if(sa[i]>=k)y[p++]=sa[i]-k;

 77         //基数排序第一关键字

 78         for(i=0;i<m;i++)c[i]=0;

 79         for(i=0;i<n;i++)c[x[y[i]]]++;

 80         for(i=1;i<m;i++)c[i]+=c[i-1];

 81         for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];

 82         //根据sa和x数组计算新的x数组

 83         swap(x,y);

 84         p=1;x[sa[0]]=0;

 85         for(i=1;i<n;i++)

 86             x[sa[i]]=y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;

 87         if(p>=n)break;   //已经排好序，直接退出

 88         m=p;     //下次基数排序的最大值

 89     }

 90 }

 91 

 92 void getHeight(int s[],int n)

 93 {

 94     int i,j,k=0;

 95     for(i=0;i<=n;i++)rank[sa[i]]=i;

 96     for(i=0;i<n;i++){

 97         if(k)k--;

 98         j=sa[rank[i]-1];

 99         while(s[i+k]==s[j+k])k++;

100         height[rank[i]]=k;

101     }

102 }

103 

104 void rmq_init(int a[])

105 {

106     int i,j;

107     for(i=1;i<=n;i++)d[i][0]=a[i];

108     for(j=1;(1<<j)<=n;j++){

109         for(i=1;i+(1<<j)-1<=n;i++){

110             d[i][j]=Min(d[i][j-1],d[i+(1<<(j-1))][j-1]);

111         }

112     }

113 }

114 

115 int rmq(int l,int r)

116 {

117     int k=0;

118     while((1<<(k+1))<=r-l+1)k++;

119     return Min(d[l][k],d[r-(1<<k)+1][k]);

120 }

121 

122 int lcp(int a,int b)

123 {

124     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

125     int ra=rank[a],rb=rank[b];

126     if(ra>rb)swap(ra,rb);

127     return rmq(ra+1,rb);

128 }

129 

130 int w[N];

131 

132 int main(){

133  //   freopen("in.txt","r",stdin);

134     int i,j,k,Q,a,b,la,lb;

135     LL ans1,ans2,t;

136     w[0]=1;

137     for(i=k=1;i<N;i=j,k++)

138         for(j=i*10;i<j && i<N;i++)w[i]=k;

139     while(~scanf("%s",s))

140     {

141         n=strlen(s);

142         for(i=0;i<n;i++)

143             num[i]=s[i]-'a'+1;

144         num[n]=0;m=27;

145         build_sa(num,n+1,m);

146         getHeight(num,n);

147         rmq_init(height);

148 

149         scanf("%d",&Q);

150         ans1=(LL)Q,ans2=(LL)2*Q;

151         scanf("%d%d",&la,&lb);

152         ans1+=(LL)lb-la;ans2+=(LL)lb-la+1;

153         while(--Q){

154             scanf("%d%d",&a,&b);

155             ans1+=(LL)b-a;

156             t=(LL)Min(lcp(la,a),lb-la,b-a);

157             ans2+=(LL)b-a-t+w[t];

158         //    printf("  %I64d %d %d %d\n",t,lcp(la,a),lb-la,b-a);

159             la=a;lb=b;

160         }

161 

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

163 

164     }

165     return 0;

166 }