牛客网暑期ACM多校训练营（第二场）I - car(思维)

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

4

备注:

题目大意：在N*N 的网格里，A 选择在边上防止汽车，汽车会沿着某一方向（行或列）前进，直到到达对面，两车如果同一时间到达同一格子会相撞，现在格子上被破坏了M个点，问，要使所有汽车都能到达对面，最多能放多少辆车。

思路：对于n为偶数个的时候，若没有被破坏的点。则可放n*2辆车。奇数时可放2*n-1辆，每次破坏一个点，则个点的行和列最开始就都能放了，所以我们标记一下，各使ans-- ，当n为奇数的时候有种特殊情况。特判即可（if(n%2==1&&x[(n/2)+1]==1&&y[(n/2+1)]==1)）

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 1e5+10;
int n,m;
int x[MAXN],y[MAXN];
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        int u,v;
        for(int i=0;i<m;i++){
            scanf("%d%d",&u,&v);
            x[u] = 1;
            y[v] = 1;
        }
        int ans ;
        if(n%2==0)  ans = n*2;
        else ans = 2*n-1;
        for(int i=1;i<=n;i++){
            if(x[i]==1) ans--;
            if(y[i]==1) ans--;
        }
        if(n%2==1&&x[(n/2)+1]==1&&y[(n/2+1)]==1)
            ans++;
        cout<<ans<<endl;
    }
    return 0;
}