LightOJ1010---Knights in Chessboard (规律题)

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#规律题

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本文探讨了一个关于在棋盘上放置骑士的经典问题，目标是在不违反骑士攻击规则的前提下，求解能够放置的最大骑士数量。文章提供了针对不同棋盘尺寸的解决方案，并通过样例输入输出展示了算法的有效性。

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

8 8

3 7

4 10

Case 1: 32

Case 2: 11

Case 3: 20

Problem Setter: Jane Alam Jan

规律题
如果 $n*m$ 是偶数且都大于2
答案是 $n * m / 2$
如果n和m都是奇数且都大于2
答案就是一半的行放 $m / 2 + 1$ 个,另一半放 $m / 2$ 个
如果n m 都小于等于2
n(m)为1,答案就是m(n)

m(n)有一个为2,那么我们可以考虑可以分出多少个2*2的格子
设有x个,答案是 $(x / 2 + x$ % $2) * 4$
剩下来可能有一个2*1的,这时如果x是奇数,不能放,如果x是偶数可以放

/*************************************************************************
    > File Name: LightOJ1010.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年06月08日 星期一 14时24分24秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

int main() {
    int t, icase = 1;
    scanf("%d", &t);
    while (t--) {
        int n, m;
        scanf("%d%d", &n, &m);
        int ans = 0;
        if (n >= 3 && m >= 3) {
            if (n * m % 2 == 0) {
                ans = n * m / 2;
            }
            else {
                int s = m / 2 + 1;
                ans += (n / 2 + 1) * s;
                ans += (n / 2) * (s - 1);
            }
        }
        else {
            if (n == 1) {
                ans = m;
            }
            else if (m == 1) {
                ans = n;
            }
            else {
                if (m == 2) {
                    swap(n, m);
                }
                if (m <= 3) {
                    ans = 4;
                }
                else {
                    int cnt = m / 2;
                    ans = (cnt / 2 + cnt % 2) * 4;
                    if (m % 2 && cnt % 2 == 0) {
                        ans += 2;
                    }
                }
            }
        }
        printf("Case %d: %d\n", icase++, ans);
    }
    return 0;
}