Codeforces 161C（分治性质）

Posted 2021-11-29 alphawa

要点

• 因为当前最大字符只有一个且两边是回文的，所以如果答案包含最大字符则一定是重合部分。
• 若不包含，则用此字符将两个区间分别断为两部分，则共有四种组合，答案一定为其中之一。

#include <cstdio>
#include <algorithm>
using namespace std;

int l1, l2, r1, r2;

int Divide(int a, int b, int c, int d, int depth) {
    if (a > b || c > d) return 0;
    int res = max(0, min(b, d) - max(a, c) + 1);
    if ((a <= c && b >= d) || (c <= a && d >= b))   return res;

    int mid = 1 << depth;
    int x[2], y[2], s[2], t[2];
    x[0] = min(a, mid), y[0] = min(b, mid - 1);
    x[1] = max(a, mid + 1) - mid, y[1] = max(b, mid) - mid;
    s[0] = min(c, mid), t[0] = min(d, mid - 1);
    s[1] = max(c, mid + 1) - mid, t[1] = max(d, mid) - mid;

    res = max(res, Divide(x[0], y[0], s[0], t[0], depth - 1));//ll
    res = max(res, Divide(x[0], y[0], s[1], t[1], depth - 1));//lr
    res = max(res, Divide(x[1], y[1], s[0], t[0], depth - 1));//rl
    res = max(res, Divide(x[1], y[1], s[1], t[1], depth - 1));//rr
    return res;
}

int main() {
    scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
    return !printf("%d\n", Divide(l1, r1, l2, r2, 30));
}