SPOJ 694&&SPOJ705: Distinct Substrings

DISUBSTR - Distinct Substrings

链接

题意：

　　询问有多少不同的子串。

思路：

　　后缀数组或者SAM。

　　首先求出后缀数组，然后从对于一个后缀，它有n-sa[i]-1个前缀，其中有height[rnk[i]]个被rnk[i]-1的后缀算了。所以再减去height[rnk[i]]即可。

代码：

换了板子。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>

using namespace std;

const int N = 1010;
char s[N];
int t1[N],t2[N],sa[N],height[N],rnk[N],c[N];
int n;

void get_sa(int m) {
    int *x = t1,*y = t2,i,p;
    for (i=1; i<=m; ++i) c[i] = 0;
    for (i=1; i<=n; ++i) x[i]=s[i],c[x[i]]++;
    for (i=1; i<=m; ++i) c[i] += c[i-1];
    for (i=n; i>=1; --i) sa[c[ x[i] ]--] = i;
    for (int k=1; k<=n; k<<=1) {
        p = 0;
        for (i=n-k+1; i<=n; ++i) y[++p] = i;
        for (i=1; i<=n; ++i) if (sa[i]>k) y[++p] = sa[i]-k;
        for (i=1; i<=m; ++i) c[i] = 0;
        for (i=1; i<=n; ++i) c[ x[y[i]] ]++;
        for (i=1; i<=m; ++i) c[i] += c[i-1];
        for (i=n; i>=1; --i) sa[c[ x[y[i]] ]--] = y[i];
        swap(x,y);
        p = 2;
        x[sa[1]] = 1;
        for (i=2; i<=n; ++i)
            x[sa[i]] = y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k] ? p-1 : p++;
        if (p > n) break;
        m = p; 
    }
}
void get_height() {
    for (int i=1; i<=n; ++i) rnk[sa[i]] = i;
    int k = 0;
    height[1] = 0;
    for (int i=1; i<=n; ++i) {
        if (rnk[i]==1) continue;
        if (k) k--;
        int j = sa[rnk[i]-1];
        while (i+k<=n && j+k<=n && s[i+k]==s[j+k])
         k++;
        height[rnk[i]] = k;
    }
}
void get_ans() {
    long long ans = 0;
    for (int i=1; i<=n; ++i) {
        ans += max(n-i+1-height[rnk[i]],0);
//      ans += max(n-sa[i]-height[i],0);
    }
    cout << ans << '
';
}
int main() {
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%s",s+1);
        n = strlen(s+1);
        get_sa(130);
        get_height();
        get_ans();
    }
    return 0;
}

原来的

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>

using namespace std;

const int N = 1010;
char s[N];
int t1[N],t2[N],sa[N],height[N],rnk[N],c[N];
int n;

void get_sa(int m) {
    int *x = t1,*y = t2;
    for (int i=0; i<m; ++i) c[i] = 0;
    for (int i=0; i<n; ++i) c[x[i] = s[i]]++;
    for (int i=1; i<m; ++i) c[i] += c[i-1];
    for (int i=n-1; i>=0; --i) sa[--c[x[i]]] = i;
    for (int k=1; k<=n; k<<=1) {
        int p = 0;
        for (int i=n-k; i<n; ++i) y[p++] = i;
        for (int i=0; i<n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k;
        for (int i=0; i<m; ++i) c[i] = 0;
        for (int i=0; i<n; ++i) c[x[y[i]]]++;
        for (int i=1; i<m; ++i) c[i] += c[i-1];
        for (int i=n-1; i>=0; --i) sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        p = 1;
        x[sa[0]] = 0;
        for (int i=1; i<n; ++i)
            x[sa[i]] = (y[sa[i-1]]==y[sa[i]] && sa[i-1]+k<n && sa[i]+k<n && 
            y[sa[i-1]+k]==y[sa[i]+k]) ? p-1 : p++;
            
        if (p >= n) break;
        m = p;
    }
}
void get_height() {
    for (int i=0; i<n; ++i) rnk[sa[i]] = i;
    int k = 0;
    height[0] = 0;
    for (int i=0; i<n; ++i) {
        if (!rnk[i]) continue;
        if (k) k--;
        int j = sa[rnk[i]-1];
        while (i+k < n && j+k < n && s[i+k]==s[j+k]) k++;
        height[rnk[i]] = k;
    }
}
void get_ans() {
    long long ans = 0;
    for (int i=0; i<n; ++i) {
        ans += max(n-i-height[rnk[i]],0);
//        ans += max(n-sa[i]-height[i],0);
    }
    cout << ans << '
';
}
int main() {
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%s",s);
        n = strlen(s);
        get_sa(130);
        get_height();
        get_ans();
    }
    return 0;
}