City Skyline

The best part of the day for Farmer John's cows is when the sun sets. They can see the skyline of the distant city. Bessie wonders how many buildings the city has. Write a program that assists the cows in calculating the minimum number of buildings in the city, given a profile of its skyline.

The city in profile is quite dull architecturally, featuring only box-shaped buildings. The skyline of a city on the horizon is somewhere between 1 and W units wide (1 <= W <= 1,000,000) and described using N (1 <= N <= 50,000) successive x and y coordinates (1 <= x <= W, 0 <= y <= 500,000), defining at what point the skyline changes to a certain height.

An example skyline could be:
..........................
.....XX.........XXX.......
.XXX.XX.......XXXXXXX.....
XXXXXXXXXX....XXXXXXXXXXXX

and would be encoded as (1,1), (2,2), (5,1), (6,3), (8,1), (11,0), (15,2), (17,3), (20,2), (22,1).

This skyline requires a minimum of 6 buildings to form; below is one possible set of six buildings whose could create the skyline above:

.......................... ..........................
.....22.........333....... .....XX.........XXX.......
.111.22.......XX333XX..... .XXX.XX.......5555555.....
X111X22XXX....XX333XXXXXXX 4444444444....5555555XXXXX

..........................
.....XX.........XXX.......
.XXX.XX.......XXXXXXX.....
XXXXXXXXXX....666666666666

Input

* Line 1: Two space separated integers: N and W

* Lines 2..N+1: Two space separated integers, the x and y coordinate of a point where the skyline changes. The x coordinates are presented in strictly increasing order, and the first x coordinate will always be 1.

Output

* Line 1: The minimum number of buildings to create the described skyline.

Sample Input

10 26
1 1
2 2
5 1
6 3
8 1
11 0
15 2
17 3
20 2
22 1

Sample Output

6

题意：用坐标的形式给出以y为高度的一些楼房的高度，求楼房最小个数。 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#define rush() int T;cin>>T;while(T--)
#define go(a) while(cin>>a)
#define ms(a,b) memset(a,b,sizeof a)
#define E 1e-8
using namespace std;
typedef __int64 ll;
const int idata=5e4+5;

ll n,m,t,_;
int i,j,k;
int cnt,num,ans;
int minn,maxx;
stack<int>stk;
priority_queue<ll,vector<ll>,less<ll> >q;
map<ll,int>mp;
ll a[idata];

int main()
{
    cin.tie(0);
    iostream::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        for(i=1;i<=n;i++){
            cin>>a[i]>>a[i];
        }

        num=0,a[n+1]=0;
        stk.push(0);
        for(i=1;i<=n+1;i++){
            //找到以栈顶为高度的楼层
            while(!stk.empty()&&stk.top()>a[i]){
                stk.pop();
                num++;
            }
            //避免相同时的重复操作
            if(a[i]!=stk.top()){
                stk.push(a[i]);
            }
        }
        cout<<num<<endl;
    }
    return 0;
}