Codeforces 689D Friends and Subsequences

发表于 分类于 Disqus：

链接

传送门

题意

给出序列a和序列b，问有多少组l、r满足\(\max \limits_{i=l}^{r}a_i=\min\limits_{i=l}^{r}b_i\)。

思路

对于左端点固定的区间，最大值随右端点的单调不降，最小值单调不增。 所以\(\max \limits_{i=l}^{r}a_i-\min\limits_{i=l}^{r}b_i\)也是单调不降的。因此可以对于每个左端点二分满足条件的右端点的区间求解。

代码

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

using namespace std;

typedef long long LL;

const int maxn = 200005;
const int maxm = 18;

int n, mm[maxn];
int Max[maxn][maxm], Min[maxn][maxm];

int rmq(int L, int R, bool x) {
  int k = mm[R - L + 1];
  return x ? max(Max[L][k], Max[R - (1 << k) + 1][k]) : min(Min[L][k], Min[R - (1 << k) + 1][k]);
}

int main() {
  int n;
  scanf("%d", &n);
  mm[0] = -1;
  for (int i = 1; i <= n; ++i) {
    mm[i] = (i & (i - 1)) ? mm[i - 1] : mm[i - 1] + 1;
    scanf("%d", &Max[i][0]);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &Min[i][0]);
  }
  for(int j = 1; j <= mm[n]; ++j) {
    for(int i = 1; i + (1 << j) - 1 <= n; ++i) {
      Max[i][j] = max(Max[i][j - 1], Max[i + (1 << (j - 1))][j - 1]);
      Min[i][j] = min(Min[i][j - 1], Min[i + (1 << (j - 1))][j - 1]);
    }
  }
  LL ans = 0;
  int L, R, mid;
  for (int i = 1; i <= n; ++i) {
    L = i, R = n;
    while (L <= R) {
      mid = (L + R) >> 1;
      if (rmq(i, mid, true) - rmq(i, mid, false) < 0) {
        L = mid + 1;
      } else {
        R = mid - 1;
      }
    }
    ans -= L;
    L = i, R = n;
    while (L <= R) {
      mid = (L + R) >> 1;
      if (rmq(i, mid, true) - rmq(i, mid, false) <= 0) {
        L = mid + 1;
      } else {
        R = mid - 1;
      }
    }
    ans += L;
  }
  printf("%lld\n", ans);
  return 0;
}
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