# [BZOJ 1660] Bad Hair Day

[题目链接]

[算法]

Sprease Table + 二分

时间复杂度 ： O(NlogN)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 80010
#define MAXLOG 20
typedef long long ll;

ll a[MAXN];
ll f[MAXN][MAXLOG];

template <typename T> inline void read(T &x)
{
ll f = 1; x = 0;
char c = getchar();
for (; !isdigit(c); c = getchar())
{
if (c == '-') f = -f;
}
for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
x *= f;
}
inline ll rmq(int x,int y)
{
int k = (int)(log(y - x + 1) / log(2.0));
return max(f[x][k],f[y - (1 << k) + 1][k]);
}

int main()
{

int n;
for (int i = 1; i <= n; i++) read(a[i]);
reverse(a + 1,a + n + 1);
for (int i = 1; i <= n; i++) f[i][0] = a[i];
for (int i = 1; i < MAXLOG; i++)
{
for (int j = 1; j + (1 << i) - 1 <= n; j++)
{
f[j][i] = max(f[j][i - 1],f[j + (1 << (i - 1))][i - 1]);
}
}
ll ans = 0;
for (int i = 1; i <= n; i++)
{
int l = 1,r = i - 1,t = -1;
while (l <= r)
{
int mid = (l + r) >> 1;
if (rmq(mid,i - 1) < a[i])
{
t = mid;
r = mid - 1;
} else l = mid + 1;
}
if (t != -1) ans += (ll)(i - t);
}
printf("%lld\n",ans);

return 0;

}

posted @ 2018-08-18 21:51  evenbao  阅读(102)  评论(0编辑  收藏  举报