畅通工程，How Many Tables——并查集模板

最新推荐文章于 2024-11-10 21:49:48 发布

原创最新推荐文章于 2024-11-10 21:49:48 发布 · 112 阅读

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本文深入探讨了畅通工程问题与HowManyTables问题，通过使用并查集算法，解决城镇道路连接与朋友聚会桌位分配的问题。文章提供了详细的算法实现与样例输入输出，适合对并查集算法感兴趣的读者。

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标题Title：畅通工程

Description

某省调查城镇交通状况，得到现有城镇道路统计表，表中列出了每条道路直接连通的城镇。省政府“畅通工程”的目标是使全省任何两个城镇间都可以实现交通（但不一定有直接的道路相连，只要互相间接通过道路可达即可）。问最少还需要建设多少条道路？

Input

测试输入包含若干测试用例。每个测试用例的第1行给出两个正整数，分别是城镇数目N ( < 1000 )和道路数目M；随后的M行对应M条道路，每行给出一对正整数，分别是该条道路直接连通的两个城镇的编号。为简单起见，城镇从1到N编号。
注意:两个城市之间可以有多条道路相通,也就是说
3 3
1 2
1 2
2 1
这种输入也是合法的
当N为0时，输入结束，该用例不被处理。

Output

对每个测试用例，在1行里输出最少还需要建设的道路数目。

Sample Input

4 2
1 3
4 3
3 3
1 2
1 3
2 3
5 2
1 2
3 5
999 0
0

Sample Output

1
0
2
998
Hint

Hint Huge input, scanf is recommended.

#include<stdio.h>
#include<queue>
#include<deque>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#define shuzu 1000+10
int n,m;
int x,y;
int fa[shuzu];
int findfa(int k)
{
    while(k!=fa[k])
        k=fa[k];
    return k;
}
void UNION()
{
    int X=findfa(x);
    int Y=findfa(y);
    if(X!=Y)
        fa[X]=Y;
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0)  break;
        scanf("%d",&m);
        for(int i=1;i<=n;i++)
            fa[i]=i;//初始化
        while(m--)
        {
            scanf("%d %d",&x,&y);
            UNION();
        }
        int sumCont=0;
        for(int i=1;i<=n;i++)
            if(fa[i]==i)
                sumCont++;
        printf("%d\n",sumCont-1);
    }
    return 0;
}

题目Title：How Many Tables
Description

Today is Ignatius’ birthday. He invites a lot of friends. Now it’s dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

Input

The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

Output

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

Sample Input

2
5 3
1 2
2 3
4 5

5 1
2 5

Sample Output

2
4

#include<stdio.h>
#include<queue>
#include<deque>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#define shuzu 1000+10
int n,m;
int x,y;
int fa[shuzu];
int findfa(int k)
{
    while(k!=fa[k])
        k=fa[k];
    return k;
}
void UNION()
{
    int X=findfa(x);
    int Y=findfa(y);
    if(X!=Y)
        fa[X]=Y;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            fa[i]=i;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            UNION();
        }
        int sumCont=0;
        for(int i=1;i<=n;i++)
            if(fa[i]==i)
                sumCont++;
        printf("%d\n",sumCont);
    }
    return 0;
}

还是比较经典的，当做是模板来用吧。