LightOJ 1010 Knights in Chessboard <贪心思维>

最新推荐文章于 2021-12-05 16:35:17 发布

leibniz_zhang

最新推荐文章于 2021-12-05 16:35:17 发布

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分类专栏： light oj 贪心思维题

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本文探讨了在给定尺寸的棋盘上放置骑士的问题，目标是找出最多能放置多少个骑士使得它们不会互相攻击。文章提供了一种算法实现思路，并附带源代码。

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Knights in Chessboard

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status uDebug

Description

Given an m x n chessboard where you want to place chess knights. You have to find the number of maximum knights that can be placed in the chessboard such that no two knights attack each other.

Those who are not familiar with chess knights, note that a chess knight can attack 8 positions in the board as shown in the picture below.

Input

Input starts with an integer T (≤ 41000), denoting the number of test cases.

Each case contains two integers m, n (1 ≤ m, n ≤ 200). Here m and n corresponds to the number of rows and the number of columns of the board respectively.

Output

For each case, print the case number and maximum number of knights that can be placed in the board considering the above restrictions.

Sample Input

8 8

3 7

4 10

Sample Output

Case 1: 32

Case 2: 11

Case 3: 20

Source

Problem Setter: Jane Alam Jan

令行<=列...

当行<3时单独考虑----

其他的把棋盘分成连续黑白两色---黑最多比白多1.

代码：

#include<cstdio>
void s(int ca)
{
    int m,n;
    scanf("%d%d",&m,&n);
    int s;
    printf("Case %d: ",ca);
    if (m>n)
    {
        s=m;m=n;n=s;
    }
    if (m==1)
    {
        printf("%d\n",n);
        return ;
    }
    else
        if (m==2)
    {
        int lp=n/4*2;
        if (n%4==1)
            lp++;
        if (n%4==2||n%4==3)
            lp+=2;
        printf("%d\n",lp*2);
    }
    else
    {
        printf("%d\n",(m*n+1)/2);
    }
}
int main()
{
    int t;scanf("%d",&t);
    for (int i=1;i<=t;i++)
        s(i);
    return 0;
}