bzoj 4556

Description

佳媛姐姐过生日的时候，她的小伙伴从某东上买了一个生日礼物。生日礼物放在一个神奇的箱子中。箱子外边写了

一个长为n的字符串s，和m个问题。佳媛姐姐必须正确回答这m个问题，才能打开箱子拿到礼物，升职加薪，出任CE

O，嫁给高富帅，走上人生巅峰。每个问题均有a,b,c,d四个参数，问你子串s[a..b]的所有子串和s[c..d]的最长公

共前缀的长度的最大值是多少？佳媛姐姐并不擅长做这样的问题，所以她向你求助，你该如何帮助她呢？

Input

输入的第一行有两个正整数n,m，分别表示字符串的长度和询问的个数。接下来一行是一个长为n的字符串。接下来

m行，每行有4个数a,b,c,d，表示询问s[a..b]的所有子串和s[c..d]的最长公共前缀的最大值。1<=n,m<=100,000,

字符串中仅有小写英文字母，a<=b,c<=d,1<=a,b,c,d<=n

Output

 对于每一次询问，输出答案。

Sample Input

5 5
aaaaa
1 1 1 5
1 5 1 1
2 3 2 3
2 4 2 3
2 3 2 4

Sample Output

1
1
2
2
2

看到这种题先反应是后缀数组和二分答案，关键是如何check

对于二分出的答案mid，从rank[c]开始向左找到使得min（height[i]-height[rank[c]]）>= mid 的最小的i，向右同理，因为要保证是在a-b之间的，所以用可持久化线段树查询一下是否存在这样的起始位置。

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 100005;const int NOD = 2000005;int sa[MAXN], rank[MAXN], w[MAXN], x[MAXN], i, j, k, n, q, m = 255, height[MAXN], tot;int f[MAXN][17], g[MAXN][17], nn, ans;int lc[NOD], len, rc[NOD], sum[NOD], root[MAXN], a, b, c, d, l, r;char s[MAXN];inline int get(){char c;while ((c = getchar()) < 48 || c > 57);int res = c - '0';while ((c = getchar()) >= 48 && c <= 57)res = res * 10 + c - '0';return res;}inline void insert(int &x, int y, int p, int q, int w){x = ++len;lc[x] = lc[y];rc[x] = rc[y];sum[x] = sum[y] + 1;if (p == q) return;int mid = p + q >> 1;if (mid >= w) insert(lc[x], lc[y], p, mid, w);else insert(rc[x], rc[y], mid + 1, q, w);}inline void find(int x, int y, int p, int q, int l, int r){if (p >= l && q <= r){ans += sum[x] - sum[y];return;}int mid = p + q >> 1;if (mid >= l) find(lc[x], lc[y], p, mid, l, r);if (mid < r) find(rc[x], rc[y], mid + 1, q, l, r);}inline int find_suffix(int x, int y){for(int i = nn; i >= 0; i --)if (x + (1 << i) <= n && f[x][i] >= y) x += 1 << i;return x;}inline int find_prefix(int x, int y){for(int i = nn; i >= 0; i --)if (x > (1 << i) && g[x][i] >= y) x -= 1 << i;return x;}inline bool check(int a, int b, int c, int mid){if (!mid) return 1;int l = find_prefix(rank[c], mid);int r = find_suffix(rank[c], mid);ans = 0;find(root[r], root[l - 1], 1, n, a, b - mid + 1);if (ans) return 1;else return 0;}int main(){cin >> n >> q;scanf("%s", s + 1);m = 255;for(i = 1; i <= n; i ++)++w[x[i] = s[i]];for(i = 2; i <= m; i ++)w[i] += w[i - 1];for(i = n; i >= 1; i --)sa[w[x[i]] --] = i;for(k = 1; k <= n; k <<= 1){int t = 0;for(i = n - k + 1; i <= n; i ++)rank[++t] = i;for(i = 1; i <= n; i ++)if (sa[i] > k) rank[++t] = sa[i] - k;for(i = 1; i <= m; i ++)w[i] = 0;for(i = 1; i <= n; i ++)w[x[i]] ++;for(i = 2; i <= m; i ++)w[i] += w[i - 1];for(i = n; i >= 1; i --)sa[w[x[rank[i]]]--] = rank[i];m = 0;for(i = 1; i <= n; i ++){int u = sa[i], v = sa[i - 1];if (x[u] != x[v] || x[u + k] != x[v + k]) rank[u] = ++m;else rank[u] = m;}if (m == n) break;for(i = 1; i <= n; i ++)swap(x[i], rank[i]);}for(i = 1; i <= n; i ++){if (tot) tot --;int j = sa[rank[i] - 1];while (s[i + tot] == s[j + tot]) tot ++;height[rank[i]]= tot;}for(i = 1; i <= n; i ++)insert(root[i], root[i - 1], 1, n, sa[i]);for(i = 1; i < n; i ++)g[i + 1][0] = f[i][0] = height[i + 1];nn = log2(n);for(j = 1; j <= nn; j ++)for(i = 1; i <= n - (1 << j); i ++)f[i][j] = min(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);for(j = 1; j <= nn; j ++)for(i = n; i > (1 << j); i --)g[i][j] = min(g[i][j - 1], g[i - (1 << j - 1)][j - 1]);while (q --){a = get(); b = get(); c = get(); d = get();if (a == c && b == d){printf("%d\n", c - a + 1);continue;}l = 0; r = min(d - c + 1, min(max(height[rank[c]], height[rank[c] + 1]), b - a + 1));while (l != r - 1){if (l == r) break;int mid = (l + r) >> 1;if (check(a, b, c, mid)) l = mid;else r = mid;}if (check(a, b, c, r)) printf("%d\n", r);else printf("%d\n", l);}}