SPOJ 694

题面

题意：

给一个字符串，求它有多少个不同的子串

多组数据。

Solution :

模板题，用所有的减去重复的即可。

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

using namespace std;

const int N = 1e6;

char str[N + 1];

namespace SA {

    int sa[N + 1], rk[N + 1], height[N + 1];

    int a[N+1], set[N+1], n;
    int fir[N+1], sec[N+1], buc[N+1], tp[N+1];

    void sufSort() {
        n = strlen(str + 1);
        copy(str + 1, str + 1 + n, set + 1);
        sort(set + 1, set + n + 1);
        int *end = unique(set + 1, set + n + 1);
        for (int i = 1; i <= n; ++ i)
            a[i] = lower_bound(set + 1, end, str[i]) - set;

        fill(buc, buc + 1 + n, 0);
        for (int i = 1; i <= n; ++ i) buc[a[i]] ++;
        for (int i = 1; i <= n; ++ i) buc[i] += buc[i - 1];
        for (int i = 1 ;i <= n; ++ i) rk[i] = buc[a[i] - 1] + 1;

        for (int t = 1; t <= n; t <<= 1) {
            copy(rk + 1, rk + 1 + n, fir + 1);//那张图
            for (int i = 1; i <= n; ++ i) sec[i] = i + t > n ? 0 : rk[i + t];

            fill(buc, buc + 1  + n, 0);//基排桶清空
            for (int i = 1; i <= n; ++ i) buc[sec[i]] ++;
            for (int i = 1; i <= n; ++ i) buc[i] += buc[i - 1];//统计比sec[i]小的数有多少个
            //tp[i]为第二关键字为第i大的二元组在排序前的位置
            //n - --buc[sec[i]]为排名(第几大), 排序前的位置自然是i.
            for (int i = 1; i <= n; ++ i) tp[n - --buc[sec[i]]] = i;

            fill(buc, buc + 1 + n, 0);
            for (int i = 1; i <= n; ++ i) buc[fir[i]] ++;
            for (int i = 1; i <= n; ++ i) buc[i] += buc[i - 1];
            //i为第二关键字为j大的二元组的位置，j递增，即保证sec[i]递增, 所以基排就是正确的。
            for (int j = 1, i; j <= n; ++ j) i = tp[j], sa[buc[fir[i]]--] = i;
            //手模，理解.

            bool unique = 1;//若关键字相同，则排名也要相同,而基排使rk不同
            for (int j = 1, i, last = 0; j <= n; ++ j) {
                i = sa[j];
                if (!last) rk[i] = 1;
                else if (sec[i] == sec[last] && fir[i] == fir[last])
                    rk[i] = rk[last], unique = 0;
                else
                    rk[i] = rk[last] + 1;
                last = i;
            }
            if (unique) break;//如果所有rk都不同，就证明排好了.
        }
    }

    void getHeight() {
        for (int i = 1, k = 0; i <= n; ++ i) {
            if (rk[i] == 1) k = 0;
            else {
                if (k > 0) k --;
                int j = sa[rk[i] - 1];
                while (j + k <= n && i + k <= n && str[i + k] == str[j + k]) k ++;
            }
            height[rk[i]] = k;
        }
    }
}using namespace SA;

int main() {
    int T; scanf("%d", &T);
    while (T --) {
        scanf("%s", str + 1);
        int n = strlen(str + 1);
        sufSort();
        getHeight();
        int ans = n * (n + 1) / 2;
        for (int i = 2; i <= n; ++ i) ans -= height[i];
        printf("%d\n", ans);
    }
}