【POJ1703】Find them, Catch them

最新推荐文章于 2020-07-21 12:35:15 发布

原创最新推荐文章于 2020-07-21 12:35:15 发布 · 197 阅读

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本文深入探讨了一种用于判断犯罪分子是否属于同一犯罪团伙的算法。通过处理一系列关于犯罪分子归属的消息，算法能够基于有限信息判断两个犯罪分子是否属于同一帮派。文章详细解释了如何维护犯罪分子间的联系，并通过路径压缩和关系更新来提高算法效率。

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Find them, Catch them?点击打开题目链接

Description

The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)

Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:

1. D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.

2. A [a] [b]
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

Output

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."

Sample Input

Sample Output

Not sure yet.

In different gangs.

In the same gang.

楼主在做这个提的时候花了很长时间，因为一直没搞明白如何压缩路径，然后每次就是TLE，搞得我很迷茫，好不容易压缩路的部分没问题了，又在更新结点与根的关系的步骤上除了维妮塔，出现了逻辑维妮塔，因为要更新这个点与祖先节点的关系，就要更新这个点与父节点之间的关系，那么首先更新父节点与祖先节点之间的关系......逐次往上，最终更新了这个点与祖先节点之间的关系。就是下面这个代码：

int search(int x)
{
	if(pre[x]==x)
	return x;
	int t=pre[x];
	//rel[x]=(rel[x]+rel[t])%2;
	pre[x]=search(pre[x]);
	rel[x]=(rel[x]+rel[t])%2;
	return pre[x];
 }

注释掉的那句是我WA那下iefa，我是先更新他与他父节点的关系，在压缩路径，这样最终这个点存的是这个点与他父节点的关系，而不是和他祖先节点的关系！

下面是完整代码：

#include<iostream>
#include<cstdio>
using namespace std;
int pre[100010];
int rel[100010];
long long k,ansa,ansb; 
int search(int x)
{
	if(pre[x]==x)
	return x;
	int t=pre[x];
	//rel[x]=(rel[x]+rel[t])%2;
	pre[x]=search(pre[x]);
	rel[x]=(rel[x]+rel[t])%2;
	return pre[x];
 } 
int Merge(int x,int y)
{
	int fx=search(x);
	int fy=search(y);
		pre[fx]=fy;
	rel[fx]=(rel[x]+1+rel[y])%2;
}
int main()
{
	char ch[2];
	int n,m,a,b,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
		{
			pre[i]=i;
			rel[i]=0;
		}
		for(int i=1;i<=m;i++)
		{
			scanf("%s%d%d",ch,&a,&b);
			if(ch[0]=='D') 
			Merge(a,b);
			else 
			{
				if(search(a)==search(b))
				{
					if(rel[a]!=rel[b])
						cout<<"In different gangs."<<endl;
					else
						cout<<"In the same gang."<<endl;
				}	
				else
					cout<<"Not sure yet."<<endl;
			} 
		}
	}
	return 0;
}