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SPOJ SUBST1(后缀数组)

2016-02-20 20:18 281 查看
(http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105595#problem/E)

题意:询问字符串s的子串个数

解法:后缀数组裸题

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <queue>
//#include <tr1/unordered_set>
//#include <tr1/unordered_map>
#include <bitset>
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define inf 1e9
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define clr(x, y) memset(x, y, sizeof x)
#define ll long long
#define ull unsigned long long
#define FOR(i,a,b) \
for(i=a;a<b?i<=b:i>=b;a<b?i++:i--)

const int maxn =  50000*2+100;

struct suffix_array{
char s[maxn];
int sa[maxn],t[maxn],t2[maxn],c[maxn];
int n;
void build_sa(int m)
{
int i,*x=t,*y=t2;
FOR(i,0,m-1) c[i]=0;
FOR(i,0,n-1) c[x[i]=s[i]]++;

FOR(i,1,m-1) c[i]+=c[i-1];
FOR(i,n-1,0) sa[--c[x[i]]]=i;
for(int k=1;k<=n;k<<=1){
int p=0;
FOR(i,n-k,n-1) y[p++]=i;
FOR(i,0,n-1) if(sa[i]>=k) y[p++]=sa[i]-k;

FOR(i,0,m-1) c[i]=0;
FOR(i,0,n-1) c[x[y[i]]]++;

FOR(i,0,m-1) c[i]+=c[i-1];
FOR(i,n-1,0) sa[--c[x[y[i]]]]=y[i];

swap(x,y);
p=1;
x[sa[0]]=0;
FOR(i,1,n-1)
x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&
y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
if(p>=n) break;
m=p;
}
}
int Rank[maxn],height[maxn];
void getHeight()
{
int k=0;
for(int i=0;i<n;i++) Rank[sa[i]]=i;
for(int i=0;i<n;i++){
if(k)k--;
int j=sa[Rank[i]-1];
while(s[i+k]==s[j+k]) k++;
height[Rank[i]]=k;
}
}
}su;

int main()
{
//freopen("input.txt","r",stdin);
//  freopen("output.txt","w",stdout);
int T;
scanf("%d",&T);
while(~scanf("%s",su.s))
{
int len = strlen(su.s);
su.n = len +1;
su.build_sa(255);
su.getHeight();

ll ans=0;

for(int i=0;i<=len;i++)
{
ans += len-su.sa[i]-su.height[i];
//          printf("%d %d\n",su.sa[i],su.height[i]);
}
printf("%lld\n",ans);
}
return 0;
}
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