|spoj 694|后缀数组|Distinct Substrings

spoj 694

给出一个字符串，求字符串中不相同的子串个数。

我们可以知道，字符串中的每个子串都是某个后缀的前缀，于是题目转化为求不相同的后缀的前缀问题。对于每一个SA[k]开始的后缀，将会增加n−SA[k]+1个后缀，而其中height[k]个是和前面的字符串的前缀是相同的。所以答案就是所有n−SA[k]+1−height[k]的总和

#include<cstdio>#include<cstring>#include<algorithm>#define ms(i, j) memset(i, j, sizeof i)#define FN2 "spoj694" using namespace std;const int MAXN = 1000 + 5;char s[MAXN];int n, m, a[MAXN], tp[MAXN], rk[MAXN], SA[MAXN], tax[MAXN], height[MAXN];bool cmp(int *f, int i, int k) {return f[SA[i]]==f[SA[i-1]]&&f[SA[i]+k]==f[SA[i-1]+k];}void build() {    for (int i=0;i<m;i++) tax[i] = 0;    for (int i=0;i<n;i++) tax[rk[i]=a[i]]++;    for (int i=1;i<m;i++) tax[i] += tax[i-1];    for (int i=n-1;i>=0;i--) SA[--tax[rk[i]]] = i;    int p;    for (int k=1;k<=n;k*=2) {        p = 0;        for (int i=n-k;i<n;i++) tp[p++] = i;        for (int i=0;i<n;i++) if (SA[i]>=k) tp[p++] = SA[i]-k;        for (int i=0;i<m;i++) tax[i] = 0;        for (int i=0;i<n;i++) tax[rk[tp[i]]]++;        for (int i=1;i<m;i++) tax[i] += tax[i-1];        for (int i=n-1;i>=0;i--) SA[--tax[rk[tp[i]]]] = tp[i];        swap(rk, tp), p = 0, rk[SA[0]] = 0;        for (int i=1;i<n;i++) rk[SA[i]] = cmp(tp, i, k) ? p : ++p;        if (++p>=n) break;        m = p;    }}void getH() {    int k = 0;    for (int i=0;i<n;i++) {        if (k) k--;        int j = SA[rk[i]-1];        while (a[i+k]==a[j+k]) k++;        height[rk[i]] = k;    }}void init() {    scanf("%s", s);    n = strlen(s) + 1;    for (int i=0;i<n-1;i++) a[i] = s[i];    m = 128, a[n-1] = 0;}void solve() {    build(), getH();    int ans = 0;    for (int i=1;i<n;i++) {        ans += (n-1) - SA[i] - height[i];    }    printf("%d\n", ans);}int main() {    #ifndef ONLINE_JUDGE    freopen(FN2".in","r",stdin);freopen("1.out","w",stdout);    #endif    int kase; scanf("%d", &kase);    while(kase--) init(), solve();    return 0;}