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Fields Division

穆风1818 2022-04-29 阅读 75

题目来源:

2021-2022 ACM-ICPC Latin American Regional Programming Contest

时间:

2022.4.28

题解:

并查集 + 思维 (连通块思想)
要保证A和B得到的田地的产量尽可能相等,所以肯定是A拿第n块田地,B拿第n-1块田地。然后剩下的田地,只要B能拿就拿(与第n-1块田相连的就能拿),剩下的给A。这样就能保证A的田地的产量尽可能大。
因为第n块田的产量比【1,n-1】的田地的产量总和还要大,第n-1块田地同理。

代码:

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
#define endl "\n"
#define inf 0x3f3f3f3f
#define debug(x) cout<<#x<<" = "<<x<<endl;
#define IOS ios::sync_with_stdio(0); cin.tie(0)
const int M = 10 + 3e5;
const int N = 10 + 3e5;

int fa[N];

int get(int x)
{
	if(fa[x] == x) return x;
	return fa[x] = get(fa[x]);
}

void merge(int u, int v)
{
	int dx = get(u);
	int dy = get(v);
	if(dx != dy)
	{
		fa[dx] = dy;
	}
}

void slove()
{
	int n, m; cin>>n>>m;
	for(int i = 1; i <= n; i++) fa[i] = i;
	for(int i = 0; i < m; i++)
	{
		int u, v; cin>>u>>v;
		if(max(u, v) == n) continue;
		merge(u, v);
	}
	for(int i = 1; i <= n; i++)
	{
		if(get(i) == get(n-1)) cout<<'B';
		else cout<<"A";
	}
	cout<<endl;
}



int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	freopen("out.out", "w", stdout);
#endif
	IOS;
	slove();
	return 0;
}
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