car

最新推荐文章于 2025-05-31 18:25:24 发布

原创最新推荐文章于 2025-05-31 18:25:24 发布 · 172 阅读

0 ·

CC 4.0 BY-SA版权

比赛同时被 2 个专栏收录

76 篇文章

订阅专栏

水题、规律、思维

25 篇文章

订阅专栏

本文介绍了一个算法问题，目标是在一个n*n的网格中放置最多的赛车，并确保它们不会因碰撞或经过损坏的网格而受损。文章详细解释了赛车的放置规则、网格的损坏情况以及如何计算可放置的最大赛车数量。

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

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld

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

备注:

题意：车只能放方框一圈，m个坏方格，位于行的车只能按行走，位于列的车只能按列走，所有的车同时出发，速度一样，车相撞会毁坏，走到坏方格也会损坏，如何放车，才能放更多的车。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,a[100005],b[100005];
int main()
{
   int i,j,x,y,sum;
   scanf("%d%d",&n,&m);
   if(m==0)
   {
       if(n%2)
        printf("%d\n",2*n-1);//推的规律
       else
        printf("%d\n",2*n);
   }
   else
   {
       memset(a,0,sizeof(a));
       memset(b,0,sizeof(b));
       for(i=1;i<=m;i++)
       {
           scanf("%d%d",&x,&y);
           a[x]=1;
           b[y]=1;
       }
       sum=0;
       for(i=1;i<=n;i++)
       {
           if(a[i]==0)
           sum++;
       }
       for(i=1;i<=n;i++)
       {
           if(b[i]==0)
            sum++;
       }
       if(n%2&&a[(n+1)/2]==0&&b[(n+1)/2]==0)//奇数的时候行和列只能放一辆车
        sum--;
       printf("%d\n",sum);
   }
}