pat 1013 Battle Over Cities(25 分) (并查集)

本文探讨了在战争环境下,城市间高速公路网络的连通性维护问题。通过算法设计,当某一城市被敌方占领导致相关道路中断时,能够迅速判断并修复必要的道路,以保持其余城市间的连接。示例展示了3个城市、2条道路的情况下,如果城市1被占领,需修复1条道路以保持城市2和城市3的连接。

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1013 Battle Over Cities(25 分)

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting city1​​-city2​​ and city1​​-city3​​. Then if city1​​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city2​​-city3​​.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing Knumbers, which represent the cities we concern.

Output Specification:

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

Sample Input:

3 2 3
1 2
1 3
1 2 3

Sample Output:

1
0
0
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <string>
 6 #include <map>
 7 #include <stack>
 8 #include <vector>
 9 #include <queue>
10 #include <set>
11 #define LL long long
12 #define INF 0x3f3f3f3f
13 using namespace std;
14 
15 const int MAXN = 1e6 + 10;
16 
17 int n, m, k, pre[MAXN], t;
18 
19 struct node
20 {
21     int a, b;
22 }edge[MAXN];
23 
24 void init()
25 {
26     for (int i = 1; i <= n; ++ i)
27         pre[i] = i;
28 }
29 
30 int my_find(int x)
31 {
32     int n = x;
33     while (n != pre[n])
34         n = pre[n];
35     int i = x, j;
36     while (n != pre[i])
37     {
38         j = pre[i];
39         pre[i] = n;
40         i = j;
41     }
42     return n;
43 }
44 
45 int main()
46 {
47     scanf("%d%d%d", &n, &m, &k);
48     for (int i = 0; i < m; ++ i)
49         scanf("%d%d", &edge[i].a, &edge[i].b);
50     for (int i = 0; i < k; ++ i)
51     {
52         init();
53         scanf("%d", &t);
54         for (int i = 0; i < m; ++ i)
55         {
56             if (edge[i].a == t || edge[i].b == t)
57                 continue;
58             int n1 = my_find(edge[i].a), n2 = my_find(edge[i].b);
59             if (n1 != n2) pre[n1] = n2;
60         }
61         int ans = 0, temp = 1 == t ? my_find(2) : my_find(1);
62         for (int i = 1; i <= n; ++ i)
63         {
64             if (i == pre[i] && i != t)
65                 ++ ans;
66         }
67 
68         printf("%d\n", -- ans);
69     }
70     return 0;
71 }

 

转载于:https://www.cnblogs.com/GetcharZp/p/9599431.html

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