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 }
