AtCoder题解——ACL Beginner Contest——C - Connect Cities

题目相关

题目链接

AtCoder ACL Beginner Contest C题，https://atcoder.jp/contests/abl/tasks/abl_c。

Problem Statement

There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City Ai and City Bi.

Snuke can perform the following operation zero or more times:

• Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.

After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).

What is the minimum number of roads he must build to achieve the goal?

Input

Input is given from Standard Input in the following format:

N M
A1 B1
:
AM BM

Output

Print the answer.

Samples1

Sample Input 1

3 1
1 2

Sample Output 1

1

Explaination

Initially, there are three cities, and there is a road between City 1 and City 2.

Snuke can achieve the goal by building one new road, for example, between City 1 and City 3. After that,

• We can travel between 1 and 2 directly.
• We can travel between 1 and 3 directly.
• We can travel between 2 and 3 by following both roads (2 - 1 - 3).

Constraints

• 2 ≤ N ≤ 100,000
• 1 ≤ M ≤ 100,000
• 1 ≤ Ai < Bi ≤N
• No two roads connect the same pair of cities.
• All values in input are integers.

题解报告

题目翻译

有 N 个城市，编号为 1 到 N。M 条双向道路，编号 1 到 M。第 i 条道路 Ai，连接了城市 Ai 和 Bi。

我们下面进行以下操作零次或者多次：

• 选择两个不同城市，而且这两个城市之间没有直接道路连接。建立一条新的道路，连接这两个城市。

当完成这些操作后，我们应该可以从任意一个城市，移动到其他任何一个城市，不管中间是否需要经过其他城市。

为了达到这个目标，我们最少需要建设几条道路。

题目分析

非常典型的一个并查集问题。我们只需要根据题目提供的数据，如果这两个城市的父节点不是一样的，我们将这两个城市建立并集，并记录下当前操作次数。假设这个数据为 cnt。

我们知道 n 个城市最少只需要 n-1 条马路，就可以互相连接。

因此最终的答案是 n-1-cnt。

AC 参考代码

//https://atcoder.jp/contests/abl/tasks/abl_c
//查并集
#include <bits/stdc++.h>

using namespace std;

map<int, int> ds;
map<int, int> ranks;

int find_root(int x) {
    return x==ds[x]?x:ds[x]=find_root(ds[x]);
}

void union_set(int x, int y) {
    int x_root=find_root(x);
    int y_root=find_root(y);
    if (x_root!=y_root) {
        if (ranks[x_root]>ranks[y_root]) {
            ds[y_root]=x_root; 
        } else if (ranks[x_root]<ranks[y_root]) {
            ds[x_root]=y_root;
        } else {
            ds[x_root]=y_root;
            ranks[y_root]++;
        }
    }
}

int main() {
    int n,m;
    cin>>n>>m;

    //初始化
    for (int i=1; i<=n; i++) {
        ds[i]=i;
        ranks[i]=0;
    }

    int cnt=0;
    for (int i=1; i<=m; i++) {
        int x,y;
        cin>>x>>y;

        x=find_root(x);
        y=find_root(y);
        if (x==y) {
            continue;
        }

        union_set(x, y);
        cnt++;
    }

    cnt=n-1-cnt;
    if (cnt<0) {
        cnt=0;
    }
    cout<<cnt<<"\n";

    return 0;
}