SPOJ - DISUBSTR 求串中子串的个数

\(height\)简单应用
#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<vector>

#include<stack>

#include<queue>

#include<set>

#include<map>

#define rep(i,j,k) for(register int i=j;i<=k;i++)

#define rrep(i,j,k) for(register int i=j;i>=k;i--)

#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])

#define iin(a) scanf("%d",&a)

#define lin(a) scanf("%lld",&a)

#define din(a) scanf("%lf",&a)

#define s0(a) scanf("%s",a)

#define s1(a) scanf("%s",a+1)

#define print(a) printf("%lld",(ll)a)

#define enter putchar('\n')

#define blank putchar(' ')

#define println(a) printf("%lld\n",(ll)a)

#define IOS ios::sync_with_stdio(0)

using namespace std;

const int maxn = 1e6+11;

const int oo = 0x3f3f3f3f;

const double eps = 1e-7;

typedef long long ll;

ll read(){

    ll x=0,f=1;register char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

char str[maxn];

struct SA{

    int Rank[maxn],sa[maxn],tsa[maxn],A[maxn],B[maxn];

    int cntA[maxn],cntB[maxn];

    int height[maxn],best[maxn][30],n;//height[i]:第sa[i]与sa[i-1]的cp

    void get(){

        n=strlen(str+1);

        rep(i,0,300) cntA[i]=0;

        rep(i,1,n) cntA[str[i]]++;

        rep(i,1,300) cntA[i]+=cntA[i-1];

        rrep(i,n,1) sa[cntA[str[i]]--]=i;

        Rank[sa[1]]=1;

        rep(i,2,n){

            if(str[sa[i]]==str[sa[i-1]]){

                Rank[sa[i]]=Rank[sa[i-1]];

            }else{

                Rank[sa[i]]=1+Rank[sa[i-1]];

            }

        }

        for(int l=1;Rank[sa[n]]<n;l<<=1){

            rep(i,1,n) cntA[i]=cntB[i]=0;

            rep(i,1,n) cntA[A[i]=Rank[i]]++;

            rep(i,1,n) cntB[B[i]=(i+l<=n?Rank[i+l]:0)]++;

            rep(i,1,n) cntA[i]+=cntA[i-1],cntB[i]+=cntB[i-1];

            rrep(i,n,1) tsa[cntB[B[i]]--]=i;

            rrep(i,n,1) sa[cntA[A[tsa[i]]]--]=tsa[i];

            Rank[sa[1]]=1;

            rep(i,2,n){

                bool flag=A[sa[i]]==A[sa[i-1]]&&B[sa[i]]==B[sa[i-1]];

                flag=!flag;

                Rank[sa[i]]=Rank[sa[i-1]]+flag;

            }

        }

    }

    void ht(){

        int j=0;

        rep(i,1,n){

            if(j) j--;

            while(str[i+j]==str[sa[Rank[i]-1]+j]) j++;

            height[Rank[i]]=j;

        }

    }

    void rmq(){

        rep(i,1,n) best[i][0]=height[i];

        for(int i=1;(1<<i)<=n;i++){

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

                best[j][i]=min(best[j][i-1],best[j+(1<<(i-1))][i-1]);

            }

        }

    }

    int query(int l,int r){

        if(l==r)return -oo;

        if(l>r)swap(l,r);

        l++;

        int k=log2(r-l+1);

        return min(best[l][k],best[r-(1<<k)+1][k]);

    }

}sa;

char str1[maxn];

int main(){

	int T=read();

    while(T--){

    	s1(str);

        int len=strlen(str+1);

        sa.get();

        sa.ht();

        ll ans=0;

		rep(i,1,len){

			ans+=len-sa.sa[i]+1-sa.height[i];

		}

        println(ans);

    }

    return 0;

}
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SPOJ - DISUBSTR 求串中子串的个数

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