并查集

最新推荐文章于 2024-07-29 20:58:50 发布

qq789045

最新推荐文章于 2024-07-29 20:58:50 发布

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本文链接：https://blog.youkuaiyun.com/qq789045/article/details/52741644

本文解析了三个POJ竞赛题目：学生病毒感染、堆方块和食物链。通过并查集算法解决不同问题，如感染传播、方块堆叠稳定性和食物链关系分析。

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POJ 1611学生病毒感染

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<string>
#include<queue>
#include<ctype.h>
#include<map>
#include<math.h>
using namespace std;

const int MAX=30000;

int parent[MAX+10];

int total[MAX+10];

int n,m;


int getParent(int a)
{
	if(parent[a]!=a)
		return parent[a]=getParent(parent[a]);
	else
		return a;
}

void merge(int a,int b)
{
	int p1=getParent(a);
	int p2=getParent(b);
	if(p1==p2)
		return;
	
	else
	{
		parent[p1]=p2;
		total[p2]+=total[p1];
	}
}

int main()
{
	while(scanf("%d%d",&n,&m)==2)
	{
		if(n==0 && m==0)
			return 0;
		
		for(int i=0;i<n;i++)
			parent[i]=i;
		
		for(int i=0;i<n;i++)
			total[i]=1;
		
		int a,b;
		
		for(int i=0;i<m;i++)
		{
			int k;
			scanf("%d",&k);
			if(k==1)
			{
				scanf("%d",&a);
			}
			else
			{
				scanf("%d",&a);
				for(int j=1;j<k;j++)
				{
					scanf("%d",&b);
					merge(a,b);
				}
			}
			
		}
		
		printf("%d\n",total[getParent(0)]);
		
		
	}
}

POJ 1988 堆方块

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<string>
#include<queue>
#include<ctype.h>
#include<map>
#include<math.h>
using namespace std;

const int MAX = 31000;

int parent[MAX];

int sum[MAX];

int under[MAX];




int getParent(int a)
{
	if (parent[a] == a)
		return a;

	int t = getParent(parent[a]);
	under[a] += under[parent[a]];
	parent[a] = t;
	return t;
}

void merge(int a, int b)
{
	int p1 = getParent(a);
	int p2 = getParent(b);
	if (p1 == p2)
		return;

	else
	{
		parent[p1] = p2;

		under[p1] += sum[p2];//?
		sum[p2] += sum[p1];
	}
}

int main()
{

	for (int i = 0; i<MAX ; i++)
	{
		parent[i] = i;
		sum[i] = 1;
		under[i] = 0;
	}

	int p;
	scanf("%d", &p);

	char s[20];
	for (int i = 0; i<p; i++)
	{
		scanf("%s", s);
		if (s[0] == 'M')
		{
			int a, b;
			scanf("%d%d", &a, &b);
			merge(a, b);
		}
		else
		{
			int a;
			scanf("%d", &a);
			getParent(a);
			printf("%d\n", under[a]);
		}
	}
}

POJ 1182 食物链

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<string>
#include<queue>
#include<ctype.h>
#include<map>
#include<math.h>
using namespace std;

const int MAXN = 50000 + 10;

int parent[MAXN];

int relation[MAXN];

int n, k;

int ans;


int getParent(int a)
{
	if (parent[a] == a)
		return a;

	int t = getParent(parent[a]);

	relation[a] = ((relation[a] + relation[parent[a]]) % 3);
	parent[a] = t;
	return t;
}


void merge(int a, int b, int p1, int p2, int d)
{
	parent[p1] = p2;
	relation[p1] = (3 - relation[a] + d - 1 + relation[b]) % 3;
}



int main()
{
	scanf("%d%d", &n, &k);

	for (int i = 1; i <= n; i++)
	{
		parent[i] = i;
		relation[i] = 0;
	}
	ans = 0;

	int d, x, y;

	for (int i = 1; i <= k; i++)
	{
		scanf("%d%d%d", &d, &x, &y);
		if (x > n || y > n)
			ans++;
		else if (d == 2 && x == y)
			ans++;
		else
		{
			int p1 = getParent(x);
			int p2 = getParent(y);

			if (p1 == p2)
			{
				if (d - 1 != (relation[x] + 3 - relation[y]) % 3)
					ans++;
			}
			else//p1!=p2
			{
				merge(x, y, p1, p2, d);
			}
		}
	}
	printf("%d\n", ans);

}