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). 给我们一个由一些矩形构造出来的图，我们需要找到最少矩形的块数来覆盖它 但是这个输入比较奇怪,针对上图,解释如下: 1.1代表在第一列,有高度为1的矩形,矩形由"X"组成.这个矩形有多宽呢,这里并没有告诉你 2.2代表在第二列,有高度为2的矩形,这就间接告诉了你,前面那个矩形有多宽,宽度即2-1 5.1代表在第五列,有高度为1的矩形,这就间接告诉了你,前面那个矩形有多宽,宽度即5-2
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
INPUT DETAILS:
The case mentioned above
1 1
2 2
5 1
6 3
8 1
11 0
15 2
17 3
20 2
22 1
INPUT DETAILS:
The case mentioned above
Sample Output
6
一直没看懂题意……orz了黄巨大之后懂了是用单调找……但是我还是不懂题意……
莫名其妙的写了个单调栈就A了
#include<cstdio>
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int zhan[500001];
int n,m,top,h,ans;
int main()
{
n=read();m=read();
for(int i=1;i<=n;i++)
{
h=read();h=read();
while (zhan[top]>h) top--;
if (zhan[top]==h) ans++;
else zhan[++top]=h;
}
printf("%d",n-ans);
return 0;
}