Lightoj 1010

1010 - Knights in Chessboard

PDF (English)

Statistics

Forum

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

白块上放马或者黑块上放马能得到最优解。注意有1和2的情况特殊讨论。

/* ***********************************************
Author        :guanjun
Created Time  :2016/6/13 20:53:50
File Name     :1010.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int n,m;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int T,ans;
    cin>>T;
    for(int t=1;t<=T;t++){
        scanf("%d%d",&n,&m);
        if(n>m)swap(n,m);
        int x,y;
        if(n&1){
            x=(n+1)/2;//第一个填 1
        }
        else x=n/2;
        if(m&1){
            y=(m+1)/2;
        }
        else y=m/2;
        ans=x*y+(n-x)*(m-y);

        if(n==1)ans=m;
        if(n==2){
             ans=m/4*4;
             if((m%4)==1)ans+=2;
             if((m%4)>=2)ans+=4;
        }
        printf("Case %d: %d
",t,max(max(n,m),ans));
    }
    return 0;
}