Educational Codeforces Round 106 (Rated for Div. 2) A. Domino on Windowsill 贪心水题

• 原题链接
  

• 测试样例

  input
  5
  1 0 1
  1 0
  1 1 1
  0 0
  3 0 0
  1 3
  4 3 1
  2 2
  5 4 3
  3 1
  output
  NO
  YES
  NO
  YES
  YES

• Note

  In the first test case, $n=1,k_1=0$ and $k_2=1.$ It means that $2\times 1$ board has black cell $(1,1)$ and white cell $(2,1)$. So, you can’t place any white domino, since there is only one white cell.
  In the second test case, the board of the same size 2×1, but both cell are white. Since w=0 and b=0, so you can place 0+0=0 dominoes on the board.
  In the third test case, board $2\times 3$, but fully colored in black (since k1=k2=0), so you can’t place any white domino.
  In the fourth test case, cells $(1,1),(1,2),(1,3)$, and $(2,1)$are white and other cells are black. You can place 2 white dominoes at positions $((1,1),(2,1))$ and $((1,2),(1,3))$ and $2$ black dominoes at positions $((1,4),(2,4))((2,2),(2,3)).$

• 解题思路
  直接贪心填充，我们要竖着填充多骨诺牌，这样才能保证空间充分利用。

• 代码

/**
* @filename:A.cbp
* @Author : pursuit
* @Blog:unique_pursuit
* @email: 2825841950@qq.com
* @Date : 2021-03-18-22.29.19
*/
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn=1e5+5;
const int mod=1e9+7;

int t,n,k1,k2;
int w,b;
void solve(){
    //简单的判断问题，k1，k2取最优。
    int minn=min(k1,k2);
    bool flag1=false,flag2=false;
    if(minn>=w){
        flag1=true;
    }
    else{
        w-=minn;
        if(w<=abs(k1-k2)/2){
            flag1=true;
        }
    }
    if(!flag1){
        cout<<"NO"<<endl;
        return;
    }
    minn=min(n-k1,n-k2);
    if(minn>=b){
        flag2=true;
    }
    else{
        b-=minn;
        if(b<=abs(n-k1-(n-k2))/2){
            flag2=true;
        }
    }
    if(flag2){
        cout<<"YES"<<endl;
    }
    else{
        cout<<"NO"<<endl;
    }
}
int main(){
    while(cin>>t){
        while(t--){
            cin>>n>>k1>>k2;
            cin>>w>>b;
            solve();
        }
    }
    return 0;
}