Codeforces Round #726 (Div. 2) B. Bad Boy(贪心)

Posted 2021-06-29 issue是fw

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容易想到两个位置选在角上是最优的

设两个点分别为$p_1,p_2$,原点为$p_3$

此时距离为$dis(p_3,p_1)+dis(p_1,p_2)+dis(p_2,p_3)$

又因为$dis(p_3,p_1)+dis(p_2+p_3)=dis(p_1,p_2)$

所以距离为$2\ast dis(p_1,p_2)$

只需要使$dis(p_1,p_2)$最远即可

这样的话,一定是选对角比较好

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9+7;
const int maxn = 3e5+10;
int n,m,x,y;
signed main()
{
	int t; cin >> t;
	while( t-- )
	{
		scanf("%d%d%d%d",&n,&m,&x,&y);
		cout << 1 << " " << m << " " << n << " " << 1 << endl;
	}
	return 0;
}

这是比赛时的傻逼代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 1e9+7;
const int maxn = 3e5+10;
int fac[maxn];
int quick(int x,int n)
{
	int ans = 1;
	for( ; n ; n>>=1,x=1ll*x*x%mod )
		if( n&1 )	ans = 1ll*ans*x%mod;
	return ans;
}
int C(int n,int m)
{
	if( m>n )	return 0;
	return 1ll*fac[n]*quick( 1ll*fac[m]*fac[n-m]%mod,mod-2 )%mod;
}
typedef pair<int,int>p;
int n,m,x,y,t,a[maxn][5];
int get(int id)
{
	int dis1 = abs( a[id][1]-x )+abs( a[id][2]-y )+abs( a[id][3]-x )+abs( a[id][4]-y );
	int dis2 = abs( a[id][1]-a[id][3] )+abs( a[id][2]-a[id][4] );
	return dis1+dis2;
}
signed main()
{
	fac[0] = 1;
	for(int i=1;i<=100000;i++)	fac[i] = 1ll*fac[i-1]*i%mod;
	int t; cin >> t;
	while( t-- )
	{
		scanf("%d%d%d%d",&n,&m,&x,&y);
		a[1][1] = 1, a[1][2] = 1, a[1][3] = n, a[1][4] = m;
		a[2][1] = 1, a[2][2] = 1, a[2][3] = 1, a[2][4] = m; 
		a[3][1] = n, a[3][2] = 1, a[3][3] = 1, a[3][4] = m;
		a[4][1] = n, a[4][2] = 1, a[4][3] = n, a[4][4] = m;
		a[5][1] = 1, a[5][2] = 1, a[5][3] = n, a[5][4] = 1;
		a[6][1] = 1, a[6][2] = m, a[6][3] = n, a[6][4] = m;
		int id = 1, ans = get( 1 );
		for(int i=2;i<=6;i++)
		{
			if( get(i)>ans )	ans = get(i), id = i;
		}
		for(int i=1;i<=4;i++)	cout << a[id][i] << " ";
		cout << endl;
	}
	return 0;
}