SPOJ DISUBSTR(后缀数组)

 

传送门:DISUBSTR

题意:给定一个字符串,求不相同的子串。

分析:对于每个sa[i]贡献n-a[i]个后缀,然后减去a[i]与a[i-1]的公共前缀height[i],则每个a[i]贡献n-sa[i]-height[i]个不同子串。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 1010;

int sa[maxn]; 
int t1[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
int s[maxn];
char str[maxn];

void build_sa(int s[], int n, int m) {
    int i, j, p, *x = t1, *y = t2;
    for (i = 0; i < m; i++) c[i] = 0;
    for (i = 0; i < n; i++) c[x[i] = s[i]]++;
    for (i = 1; i < m; i++) c[i] += c[i-1];
    for (i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;

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

        swap(x, y);
        p = 1, x[sa[0]] = 0;
        for (i = 1; i < n; i++) 
            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;

        if (p >= n) break;
        m = p;
    }
}

void getHeight(int s[],int n) {
    int i, j, k = 0;
    for (i = 0; i <= n; i++)
        rank[sa[i]] = i;
    for (i = 0; i < n; i++) {
        if (k) k--;
        j = sa[rank[i]-1];
        while (s[i+k] == s[j+k]) k++;
        height[rank[i]]=k;
    }
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%s", str);
        int n = strlen(str);
        for (int i = 0; i <= n; i++)
            s[i] = str[i];
        build_sa(s, n+1, 128);
        getHeight(s, n);
        int ans = 0;
        for (int i = 1; i <= n; i++)
            ans += n - sa[i] - height[i];
        printf("%d\n", ans);
    }
    return 0;
}
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posted on 2015-03-10 19:12  lienus  阅读(181)  评论(0编辑  收藏  举报

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