BZOJ1628: [Usaco2007 Demo]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). 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

* 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.

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

 1 #include<cstdio>
 2 
 3 #include<cstdlib>
 4 
 5 #include<cmath>
 6 
 7 #include<cstring>
 8 
 9 #include<algorithm>
10 
11 #include<iostream>
12 
13 #include<vector>
14 
15 #include<map>
16 
17 #include<set>
18 
19 #include<queue>
20 
21 #include<string>
22 
23 #define inf 1000000000
24 
25 #define maxn 50000+100
26 
27 #define maxm 500+100
28 
29 #define eps 1e-10
30 
31 #define ll long long
32 
33 #define pa pair<int,int>
34 
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36 
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38 
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40 
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42 
43 using namespace std;
44 
45 inline int read()
46 
47 {
48 
49     int x=0,f=1;char ch=getchar();
50 
51     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
52 
53     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
54 
55     return x*f;
56 
57 }
58 int a[maxn],sta[maxn],n,m;
59 
60 int main()
61 
62 {
63 
64     freopen("input.txt","r",stdin);
65 
66     freopen("output.txt","w",stdout);
67 
68     n=read();m=read();
69     for1(i,n)a[i]=read(),a[i]=read();
70     int top=0,ans=n;
71     for1(i,n)
72     {
73         while(a[sta[top]]>a[i])top--;
74         if(a[sta[top]]==a[i])ans--;else sta[++top]=i;
75     }
76     printf("%d\n",ans);    
77 
78     return 0;
79 
80 }