牛课第二次多校I

本文探讨了一个受损的n*n棋盘上,如何放置汽车从边到边移动而不损坏的问题。介绍了通过排除已损毁行列来计算最多能放置多少辆汽车的方法。

链接:https://www.nowcoder.com/acm/contest/140/I
来源:牛客网

题目描述

White Cloud has a square of n*n from (1,1) to (n,n).
White Rabbit wants to put in several cars. Each car will start moving at the same time and move from one side of one row or one line to the other. All cars have the same speed. If two cars arrive at the same time and the same position in a grid or meet in a straight line, both cars will be damaged.
White Cloud will destroy the square m times. In each step White Cloud will destroy one grid of the square(It will break all m grids before cars start).Any car will break when it enters a damaged grid.
White Rabbit wants to know the maximum number of cars that can be put into to ensure that there is a way that allows all cars to perform their entire journey without damage.
(update: all cars should start at the edge of the square and go towards another side, cars which start at the corner can choose either of the two directions)

For example, in a 5*5 square


legal

illegal(These two cars will collide at (4,4))


illegal (One car will go into a damaged grid)

输入描述:

The first line of input contains two integers n and m(n <= 100000,m <= 100000)
For the next m lines,each line contains two integers x,y(1 <= x,y <= n), denoting the grid which is damaged by White Cloud.

输出描述:

Print a number,denoting the maximum number of cars White Rabbit can put into.
示例1

输入

复制
2 0

输出

复制
4

备注:

 题解

  如果没有被破坏的点 ans=n*2-(n%2);
  如果有则减去相应的被破坏的行和列,但是奇数行的最中间的那个只影响一个,(此处可以考虑n==3的,最中间的点之影响一个车的填入)
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN=1e5+5;
 4 int a[MAXN],b[MAXN];
 5 int main()
 6 {
 7     int ans=0;
 8     int n,m;
 9     scanf("%d%d",&n,&m);
10     int x,y;
11     ans=n*2-(n%2);
12     for(int i=0;i<m;i++)
13     {
14         scanf("%d%d",&x,&y);
15         a[x]=1;
16         b[y]=1;
17     }
18     int k1=0,k2=0;
19     for(int i=1;i<n;i++)
20     {
21         if(a[i]==1 ) ans--;
22         if(b[i]==1) ans--;
23     }
24     if(n%2==1&&a[(n+1)/2]==1&&b[(n+1)/2]==1) ans++;
25     printf("%d\n",ans);
26     return 0;
27 }
View Code

 

转载于:https://www.cnblogs.com/-xiangyang/p/9348387.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值