[USACO06NOV]糟糕的一天Bad Hair Day BZOJ 1660 单调栈

Input
* Line 1: 牛的数量 N。 * Lines 2..N+1: 第 i+1 是一个整数，表示第i头牛的高度。
Output
* Line 1: 一个整数表示c[1] 至 c[N]的和。
Sample Input

6
10
3
7
4
12
2


Sample Output
5我们用单调栈去维护它；读入完毕后，我们从后往前遍历；用 sum 去维护一个后缀和；这是基于这样的一个思想，如果我在你后面而且我比你高，那么你能看见的我也能看见；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}

ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/

ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}

int a[maxn];
int sum[maxn];
ll ans;
int sk[maxn];
int main()
{
//ios::sync_with_stdio(0);
int n; rdint(n); int top = 0;
for (int i = 1; i <= n; i++)rdint(a[i]);
for (int i = n; i >= 1; i--) {
while (top&&a[i] > a[sk[top]])sum[i] += sum[sk[top]] + 1, top--;
sk[++top] = i;
}
for (int i = 1; i <= n; i++)ans +=(ll) sum[i];
cout << ans << endl;
return 0;
}


posted @ 2018-11-18 09:13  NKDEWSM  阅读(159)  评论(0编辑  收藏  举报