BZOJ1628: [Usaco2007 Demo]City skyline

1628: [Usaco2007 Demo]City skyline

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 256  Solved: 210
[Submit][Status]

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

Sample Output

6

HINT

Source

Silver

题解：

cf原题？完全忘了。。。

比较巧妙。首先应该知道 ans<=n，然后考虑ans<n什么情况下会出现？

显然应该是  有两块相同高度的积木，并且这两块积木中间没有比它们低的积木！

那就是单调栈了。

发现OI真是有时候不是看你会不会什么算法，而是你能不能想到某个算法。

代码：

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define inf 1000000000

#define maxn 50000+100

#define maxm 500+100

#define eps 1e-10

#define ll long long

#define pa pair<int,int>

#define for0(i,n) for(int i=0;i<=(n);i++)

#define for1(i,n) for(int i=1;i<=(n);i++)

#define for2(i,x,y) for(int i=(x);i<=(y);i++)

#define for3(i,x,y) for(int i=(x);i>=(y);i--)

using namespace std;

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=10*x+ch-'0';ch=getchar();}

    return x*f;

}
int a[maxn],sta[maxn],n,m;

int main()

{

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

    n=read();m=read();
    for1(i,n)a[i]=read(),a[i]=read();
    int top=0,ans=n;
    for1(i,n)
    {
        while(a[sta[top]]>a[i])top--;
        if(a[sta[top]]==a[i])ans--;else sta[++top]=i;
    }
    printf("%d\n",ans);    

    return 0;

}