【bzoj3238】 Ahoi2013—差异

http://www.lydsy.com/JudgeOnline/problem.php?id=3238 (题目链接)

题意

　　给出一个字符串，求${\sum_{1<=i<j<=n} len(T_i)+len(T_j)-2*lcp(T_i,T_j)}$，其中${T_i}$表示i的后缀

Solution

　　很简单，求个后缀数组，然后单调栈扫一遍就可以了。

细节

　　开LL

代码

// bzoj3238
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<ctime>
#define LL long long
#define inf 1<<30
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=500010;
int sa[maxn],rank[maxn],height[maxn],n;
char s[maxn];

struct data {LL w,c;}st[maxn];
namespace Suffix {
int wa[maxn],wb[maxn],ww[maxn];
bool cmp(int *r,int a,int b,int l) {
return r[a]==r[b] && r[a+l]==r[b+l];
}
void da(char *r,int *sa,int n,int m) {
int i,j,p,*x=wa,*y=wb;
for (i=0;i<=m;i++) ww[i]=0;
for (i=1;i<=n;i++) ww[x[i]=r[i]]++;
for (i=1;i<=m;i++) ww[i]+=ww[i-1];
for (i=n;i>=1;i--) sa[ww[x[i]]--]=i;
for (p=0,j=1;p<n;j*=2,m=p) {
for (p=0,i=n-j+1;i<=n;i++) y[++p]=i;
for (i=1;i<=n;i++) if (sa[i]>j) y[++p]=sa[i]-j;
for (i=0;i<=m;i++) ww[i]=0;
for (i=1;i<=n;i++) ww[x[y[i]]]++;
for (i=1;i<=m;i++) ww[i]+=ww[i-1];
for (i=n;i>=1;i--) sa[ww[x[y[i]]]--]=y[i];
for (swap(x,y),p=x[sa[1]]=1,i=2;i<=n;i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j) ? p : ++p;
}
}
void calheight(char *r,int *sa,int n) {
for (int i=1;i<=n;i++) rank[sa[i]]=i;
for (int k=0,i=1;i<=n;i++) {
if (k) k--;
int j=sa[rank[i]-1];
while (r[i+k]==r[j+k]) k++;
height[rank[i]]=k;
}
}
}

int main() {
scanf("%s",s+1);
int n=strlen(s+1);
Suffix::da(s,sa,n,300);
Suffix::calheight(s,sa,n);
LL S=0,ans=0;int top=0;
for (int i=1;i<=n;i++) ans+=(LL)(n-1)*i;
for (int i=1;i<=n;i++) {
ans-=S*2;
while (top>1 && height[i+1]<st[top-1].w) {
S-=(LL)(st[top].w-st[top-1].w)*st[top].c;
st[top-1].c+=st[top].c;
st[top--]=(data){0,0};
}
if (height[i+1]<st[top].w) {
S-=(LL)(st[top].w-height[i+1])*st[top].c;
st[top].w=height[i+1];
}
if (st[top].w==height[i+1]) st[top].c++,S+=height[i+1];
else st[++top]=(data){height[i+1],1},S+=height[i+1];
}
printf("%lld",ans);
return 0;
}