bzoj3238 [Ahoi2013]差异

Posted 2020-06-17

题目链接

树状数组+单调栈

计算出每个后缀的前面、后面第一个h[]比它小的(前闭后开)，乘起来计算答案

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<string>
 7 #include<cmath>
 8 #include<ctime>
 9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<set>
13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
15 #define Clear(a,b) memset(a,b,sizeof(a))
16 #define inout(x) printf("%d",(x))
17 #define douin(x) scanf("%lf",&x)
18 #define strin(x) scanf("%s",(x))
19 #define LLin(x) scanf("%lld",&x)
20 #define op operator
21 #define CSC main
22 typedef unsigned long long ULL;
23 typedef const int cint;
24 typedef long long LL;
25 using namespace std;
26 void inin(int &ret)
27 {
28     ret=0;int f=0;char ch=getchar();
29     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=1;ch=getchar();}
30     while(ch>=‘0‘&&ch<=‘9‘)ret*=10,ret+=ch-‘0‘,ch=getchar();
31     ret=f?-ret:ret;
32 }
33 char s[500050];
34 int n,sa[500050],c[500050],t[500050],t2[500050],h[500050];
35 void build_sa(int m)
36 {
37     int *x=t,*y=t2;
38     re(i,0,n-1)x[i]=s[i],c[x[i]]++;
39     re(i,1,m)c[i]+=c[i-1];
40     rre(i,n-1,0)sa[--c[x[i]]]=i;
41     for(int k=1;k<=n;k<<=1)
42     {
43         int p=0;
44         rre(i,n-1,n-k)y[p++]=i;
45         re(i,0,n-1)if(sa[i]>=k)y[p++]=sa[i]-k;
46         re(i,0,m-1)c[i]=0;
47         re(i,0,n-1)c[x[y[i]]]++;
48         re(i,1,m-1)c[i]+=c[i-1];
49         rre(i,n-1,0)sa[--c[x[y[i]]]]=y[i];
50         swap(x,y);
51         x[sa[0]]=0,p=1;
52         re(i,0,n-1)x[sa[i]]=y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++;
53         if(p>=n)return ;
54         m=p;
55     }
56 }
57 int rank[500050];
58 void build_height()
59 {
60     int k=0;
61     re(i,0,n-1)rank[sa[i]]=i;
62     re(i,0,n-1)
63     {
64         if(k)k--;
65         if(!rank[i])continue;
66         int j=sa[rank[i]-1];
67         while(s[i+k]==s[j+k])k++;
68         h[rank[i]]=k;
69     }
70 }
71 int l[500050],r[500050],sta[500050];
72 LL ans;
73 int main()
74 {
75     strin(s);
76     n=strlen(s);
77     build_sa(256);
78     build_height();
79     int top=0;
80     re(i,0,n-1)
81     {
82         while(top&&h[sta[top]]>=h[i])top--;
83         l[i]=(top?sta[top]+1:0);
84         sta[++top]=i;
85     }
86     top=0;
87     rre(i,n-1,0)
88     {
89         while(top&&h[sta[top]]>h[i])top--;
90         r[i]=(top?sta[top]-1:n-1);
91         sta[++top]=i;
92     }
93     ans+=(LL)(1+n)*n/2*(n-1);
94     re(i,0,n-1)
95         ans-=2LL*(i-l[i]+1)*(r[i]-i+1)*h[i];
96     printf("%lld",ans);
97      return 0;
98 }