L2-013 红色警报 (25分)【搜索 或 并查集】

L2-013 红色警报 (25分)

战争中保持各个城市间的连通性非常重要。本题要求你编写一个报警程序，当失去一个城市导致国家被分裂为多个无法连通的区域时，就发出红色警报。注意：若该国本来就不完全连通，是分裂的k个区域，而失去一个城市并不改变其他城市之间的连通性，则不要发出警报。

输入格式：

输入在第一行给出两个整数N（0 < N ≤ 500）和M（≤ 5000），分别为城市个数（于是默认城市从0到N-1编号）和连接两城市的通路条数。随后M行，每行给出一条通路所连接的两个城市的编号，其间以1个空格分隔。在城市信息之后给出被攻占的信息，即一个正整数K和随后的K个被攻占的城市的编号。

注意：输入保证给出的被攻占的城市编号都是合法的且无重复，但并不保证给出的通路没有重复。

输出格式：

对每个被攻占的城市，如果它会改变整个国家的连通性，则输出Red Alert: City k is lost!，其中k是该城市的编号；否则只输出City k is lost.即可。如果该国失去了最后一个城市，则增加一行输出Game Over.。

输入样例：

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

输出样例：

City 1 is lost.
City 2 is lost.
Red Alert: City 0 is lost!
City 4 is lost.
City 3 is lost.
Game Over.

出自于： https://blog.youkuaiyun.com/qq_40946921/article/details/88091058

 方法一：搜索

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
using namespace std;
vector<int> v[505];
int book[505];
int n, m, k, u;
void dfs(int pos)
{
    book[pos] = 1;
    int t = v[pos].size();
    for (int i = 0; i < t; i++)
        if (book[v[pos][i]] == 0)
            dfs(v[pos][i]);
}
int cnt()
{
    int count = 0;
    for (int i = 0; i < n; i++)
    {
        if (book[i] == 0)
        {
            count++;
            dfs(i);
        }
    }
    memset(book, 0, sizeof(book));
    return count;
}
int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int sum = cnt();
    vector<int> lost;
    cin >> k;

    while (k--)
    {
        cin >> u;
        lost.push_back(u);
        for (int i = 0; i < lost.size(); i++)
            book[lost[i]] = 1;

        int cur = cnt();
        if (cur <= sum)
            printf("City %d is lost.\n", u);
        else if (cur > sum)
            printf("Red Alert: City %d is lost!\n", u);

        sum = cur;
    }
    if (n == lost.size())
        printf("Game Over.\n");

    return 0;
}

 方法二：并查集

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
int n, m, k, u;
int father[505], vis[505];
struct node
{
    int a, b;
} road[5005];
int count() //通过并查集查找连通分支数
{
    int count = 0;
    for (int i = 0; i < n; i++)
        if (father[i] == i && vis[i] == 0)
            count++;
    return count;
}
void init()
{
    for (int i = 0; i < 505; i++)
        father[i] = i;
}
int find(int x)
{
    int t = x;
    while (t != father[t])
    {
        t = father[t];
    }
    return t;
}
void Union(int x, int y)
{
    int fx = find(x);
    int fy = find(y);
    if (fx != fy)
        father[fy] = fx;
}
int main()
{
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n >> m;
    init();
    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        road[i].a = a;
        road[i].b = b;
        Union(a, b);
    }
    int sum = count();
    cin >> k;
    for (int t = 0; t < k; t++)
    {
        cin >> u;
        vis[u] = 1;
        init();
        for (int i = 0; i < m; i++)
        {
            int a = road[i].a;
            int b = road[i].b;
            if (vis[a] == 0 && vis[b] == 0)
                Union(a, b);
        }

        int cur = count();
        if (cur <= sum)
            printf("City %d is lost.\n", u);
        else
            printf("Red Alert: City %d is lost!\n", u);

        sum = cur;
    }
    if (n == k)
        printf("Game Over.\n");

    return 0;
}