City Skyline --（单调队列）

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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

Hint

INPUT DETAILS:
The case mentioned above

题意：给出一个轮廓，问组成轮廓的最少楼房个数

很有意思单调队列，创建严格递增单调队列，每有元素出列就说明确定了一个楼房

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define pi acos(-1.0)#define inf 0x3f3f3f#define M 50005int n,w,q[M],x,y[M];int main(){    int i;    scanf("%d%d",&n,&w);    for(i=1;i<=n;i++)        scanf("%d%d",&x,&y[i]);    y[0]=y[n+1]=0;    int head=1,tail=0,num=0; //队尾加入0让说有元素出队    for(i=1;i<=n+1;i++)    {        while(head<=tail&&y[q[tail]]>y[i]) {tail--;num++;}        if(y[i]!=y[q[tail]]) q[++tail]=i;  //等于时不入队    }    printf("%d",num);    return 0;}

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