poj3250

maksyuki 发表于 oj 分类，标签: 数据结构-栈/单调

15 9月 2015

Bad Hair Day

Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads. Each cow i has a specified height h_i (1 ≤ h_i≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.

Consider this example:

=
=       =
=   -   =         Cows facing right -->
=   =   =
= - = = =
= = = = = =
1 2 3 4 5 6

Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.

Output

Line 1: A single integer that is the sum of c₁ through c_N.

Sample Input

Sample Output

Source

USACO 2006 November Silver

题目类型：单调栈

算法分析：本题可以转换为求所有的牛能被多少在其左边的牛看到，可以维护一个单调递减栈(严格递减)，然后每次更新完栈后，累加栈中元素的个数- 1即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const int MOD = 32767;

const int maxn = 80000 + 66;;

int ans[maxn];

int main()
{
//	ifstream cin ("aaa.txt");

	int n;
	while (cin >> n)
	{
        int i;
        for (i = 0; i < n; i++)
			cin >> ans[i];

        long long sum = 0;
        stack <int> sta;

        for (i = 0; i < n; i++)
		{
			while (!sta.empty () && sta.top () <= ans[i])
				sta.pop ();

			sta.push (ans[i]);
			sum += (sta.size () - 1);
		}

		cout << sum << endl;
	}

return 0;
}

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const int MOD = 32767;

const int maxn = 80000 + 66;;

int ans[maxn];

int main()

{

// ifstream cin ("aaa.txt");

int n;

while (cin >> n)

{

int i;

for (i = 0; i < n; i++)

cin >> ans[i];

long long sum = 0;

stack <int> sta;

for (i = 0; i < n; i++)

{

while (!sta.empty () && sta.top () <= ans[i])

sta.pop ();

sta.push (ans[i]);

sum += (sta.size () - 1);

}

cout << sum << endl;

}

return 0;

}

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