lightoj 1010 - Knights in Chessboard 【数学思维】

1010 - Knights in Chessboard

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Time Limit: 1 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
3 8 8 3 7 4 10	Case 1: 32 Case 2: 11 Case 3: 20

 

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PROBLEM SETTER: JANE ALAM JAN

题意：给定一个n*m的棋盘，现在给出马能攻击的8个位置，问最多放置多少个马使得任意两匹马之间不能相互攻击。

40000组测试数据 + 200 * 200的棋盘，匹配 状压 网络流 都不行吧。

那只能找规律了。

思路：

(1) n == 1 || m == 1时，结果显然就是max(n, m)；

(2) n > 2 && m > 2时，结果为n * m % 2 == 0 ? n*m/2 : n*m/2+1；

(3) n == 2 || m == 2时，这里我就贪心的考虑了一下，没想到过了（假设n < m）

    考虑 n == 2 && m == 1时，结果为2，n == 2 && 1 < m <= 4时结果为4。我们 是否可以贪心的认为这就是最优     解，然后把原棋盘 分成若干个n == 2 && m == 4的棋盘 以及一个n == 2 && 1 <= m <= 3的棋盘？没错就是这样     最优情况下n == 2 && m == 4放4个，n == 2 && m == 1放2个，n == 2 && 1 < m <= 3放4个。统计下就好了   

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN 500000+10
#define MAXM 50000000
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
using namespace std;
int main()
{
    int t, kcase = 1;
    Ri(t);
    W(t)
    {
        int n, m;
        Ri(n), Ri(m);
        int ans;
        if(n == 1 || m == 1)
            ans = max(n, m);
        else if(n == 2 || m == 2)
        {
            int temp = max(n, m) % 4;
            ans = max(n, m);
            if(temp)
            {
                ans = max(n, m) / 4 * 4;
                if(temp == 1)
                    ans += 2;
                else
                    ans += 4;
            }
        }
        else
            ans = n*m % 2 == 0 ? n*m/2 : n*m/2+1;
        printf("Case %d: %d\n", kcase++, ans);
    }
    return 0;
}