POJ 3250 Bad Hair Day【单调栈入门】
Bad Hair Day
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 24112 | Accepted: 8208 |
Description
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 hi (1 ≤ hi ≤ 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 ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.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 cN.
Sample Input
6
10
3
7
4
12
2
Sample Output
5
题解
单调栈入门题
代码
#include<iostream>
#include<cstdio> //EOF,NULL
#include<cstring> //memset
#include<cstdlib> //rand,srand,system,itoa(int),atoi(char[]),atof(),malloc
#include<cmath> //ceil,floor,exp,log(e),log10(10),hypot(sqrt(x^2+y^2)),cbrt(sqrt(x^2+y^2+z^2))
#include<algorithm> //fill,reverse,next_permutation,__gcd,
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#include<iterator>
#include<iomanip> //setw(set_min_width),setfill(char),setprecision(n),fixed,
#include<functional>
#include<map>
#include<set>
#include<limits.h> //INT_MAX
#include<cmath> // bitset<?> n
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(y))
#define all(x) x.begin(),x.end()
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
#define N 100001
typedef pair<int,int> P;
typedef long long LL;
typedef long long ll;
const double eps=1e-;
const double PI = acos(1.0);
const int INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+;
const int MAXN = 1e6+;
const int maxm = ;
const int maxn = + ;
const ll MOD = ;
int a[maxn];
int st[maxn];
ll ans;
int n; int main()
{
while(read(n) != EOF){
memset(st,);
for (int i = ; i < n; i++)
read(a[i]);
a[n] = inf ;
ans = ;
int top = ;
for (int i = ; i <= n; i++){
if (top == || a[i] < a[st[top - ] ])
st[top++] = i;
else{
while (top >= && a[i] >= a[st[top - ]]){
ans += (i - (st[top - ] + ) );
top--;
}
st[top++] = i;
}
}
printf("%lld\n", ans);
}
return ;
}
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