HDU6194 后缀数组

简略题意：问在一个字符串中出现次数等于k次的子串有多少种。

考虑按k对height进行分组，则当前分组对答案的贡献为:
组内所有元素的的max(0, LCP - max(组内第一个元素与之前元素的LCP, 组内最后一个元素与之后元素的LCP))。
举例来看就是：
k=2
*1. abc
*2. abcabc
*3. abcabcabc
*4. abcabcabcabc
对于分组一(*1,*2)， 其LCP为3，abc在这个组里作为前缀出现了2次，但由于*2和*3的LCP为6， 因此abc还会在*3内出现，abc出现的次数就超过了2次，当前分组的LCP对答案无贡献。
分组二同理。
而对于分组三(*3,*4)，其LCP为9，说明abcabcabc在当前分组出现了2次，由于*2和*3的LCP为6，因此abcabc出现次数超过了2次，这部分对答案无贡献。因此分组三对答案的贡献为9−6=3，出现次数为2次的子串有abcabca, abcabcab, abcabcabc。

需要特别处理的是k=1的情况，不过想法和其他情况同理。

#include <bits/stdc++.h>using namespace std;typedef long long LL;const LL N = 110000;LL n;char str[N];namespace SA {    LL sa[N], rank[N], height[N], s[N<<1], t[N<<1], p[N], cnt[N], cur[N];    LL MIN[N][30];    #define pushS(x) sa[cur[s[x]]--] = x    #define pushL(x) sa[cur[s[x]]++] = x    #define inducedSort(v) fill_n(sa, n, -1); fill_n(cnt, m, 0);                  \        for (LL i = 0; i < n; i++) cnt[s[i]]++;                                  \        for (LL i = 1; i < m; i++) cnt[i] += cnt[i-1];                           \        for (LL i = 0; i < m; i++) cur[i] = cnt[i]-1;                            \        for (LL i = n1-1; ~i; i--) pushS(v[i]);                                  \        for (LL i = 1; i < m; i++) cur[i] = cnt[i-1];                            \        for (LL i = 0; i < n; i++) if (sa[i] > 0 &&  t[sa[i]-1]) pushL(sa[i]-1); \        for (LL i = 0; i < m; i++) cur[i] = cnt[i]-1;                            \        for (LL i = n-1;  ~i; i--) if (sa[i] > 0 && !t[sa[i]-1]) pushS(sa[i]-1)    void sais(LL n, LL m, LL *s, LL *t, LL *p) {        LL n1 = t[n-1] = 0, ch = rank[0] = -1, *s1 = s+n;        for (LL i = n-2; ~i; i--) t[i] = s[i] == s[i+1] ? t[i+1] : s[i] > s[i+1];        for (LL i = 1; i < n; i++) rank[i] = t[i-1] && !t[i] ? (p[n1] = i, n1++) : -1;        inducedSort(p);        for (LL i = 0, x, y; i < n; i++) if (~(x = rank[sa[i]])) {            if (ch < 1 || p[x+1] - p[x] != p[y+1] - p[y]) ch++;            else for (LL j = p[x], k = p[y]; j <= p[x+1]; j++, k++)                if ((s[j]<<1|t[j]) != (s[k]<<1|t[k])) {ch++; break;}            s1[y = x] = ch;        }        if (ch+1 < n1) sais(n1, ch+1, s1, t+n, p+n1);        else for (LL i = 0; i < n1; i++) sa[s1[i]] = i;        for (LL i = 0; i < n1; i++) s1[i] = p[sa[i]];        inducedSort(s1);    }    template<typename T>    LL mapCharToLL(LL n, const T *str) {        LL m = *max_element(str, str+n);        fill_n(rank, m+1, 0);        for (LL i = 0; i < n; i++) rank[str[i]] = 1;        for (LL i = 0; i < m; i++) rank[i+1] += rank[i];        for (LL i = 0; i < n; i++) s[i] = rank[str[i]] - 1;        return rank[m];    }    template<typename T>    void suffixArray(LL n, const T *str) {        LL m = mapCharToLL(++n, str);        sais(n, m, s, t, p);        for (LL i = 0; i < n; i++) rank[sa[i]] = i;        for (LL i = 0, h = height[0] = 0; i < n-1; i++) {            LL j = sa[rank[i]-1];            while (i+h < n && j+h < n && s[i+h] == s[j+h]) h++;            if (height[rank[i]] = h) h--;        }    }    void RMQ_init(){        for(LL i=0; i<n; i++) MIN[i][0] = height[i+1];        for(LL j=1; (1<<j)<=n; j++){            for(LL i=0; i+(1<<j)<=n; i++){                MIN[i][j] = min(MIN[i][j-1], MIN[i+(1<<(j-1))][j-1]);            }        }    }    LL RMQ(LL L, LL R){        LL k = 0;        while((1<<(k+1)) <= R-L+1) k++;        return min(MIN[L][k], MIN[R-(1<<k)+1][k]);    }    LL LCP(LL i, LL j){        return RMQ(i, j-1);    }    void init(char *str){        str[n] = 0;        suffixArray(n, str);        RMQ_init();    }    void solve(LL k) {        LL ans = 0;        height[n+1] = 0;        if(k == 1) {            for(LL i = 1; i <= n; i++)                ans += max(0LL, n - sa[i] - max(height[i], height[i+1]));        } else {            for(LL i = 1; i + k - 1 <= n; i ++) {                ans += max(0LL, LCP(i, i+k-1) - max(height[i], height[i+k]));            }        }        printf("%lld\n", ans);    }};LL t;LL k;int main() {    scanf("%lld", &t);    while(t--) {        scanf("%lld", &k);        scanf("%s", str);        n = strlen(str);        SA::init(str);        SA::solve(k);    }    return 0;}