City Skyline（单调队列）

City Skyline

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2969 Accepted: 1367

Description

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

10 261 12 25 16 38 111 015 217 320 222 1

Sample Output

6

Hint

INPUT DETAILS: 
The case mentioned above

Source

USACO 2005 November Silver

题意：以坐标的形式给出一张图，表示一些楼房的正视图，求出楼房的最少个数。

思路:维护一个递增的单调队列,如果当队尾元素大于要入队的元素,那么此时该队尾元素可能延伸的位置就已经确定了,因为比它低的点的正视图 不能再被该队尾元素覆盖 否则其坐标不会比其低,然后每tail--一次，就增加一个楼;

代码：

#include<cstdio>#include<cmath>#include<string>#include<cstring>#include<queue>#include<algorithm>#include<iostream>#include<cstring>#define maxn 50010using namespace std;int y[maxn],n,w,x;int que[maxn];int main(){    ios::sync_with_stdio(false);    int tail=0;    cin>>n>>w;    for(int i=1;i<=n;i++){cin>>x>>y[i];}    int ans=0;    y[0]=y[n+1]=0;    for(int i=1;i<=n+1;i++)  //注意，这里是到n+1。。。wa了好几次。。。    {        while(tail && y[que[tail]] > y[i]) { tail--; ans++; }        if(y[que[tail]]!=y[i])        {            tail++;            que[tail]=i;        }    }    cout<<ans<<endl;    return 0;}