zoukankan      html  css  js  c++  java
  • pat 1013 Battle Over Cities(25 分) (并查集)

    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
    ", -- ans);
    69     }
    70     return 0;
    71 }
  • 相关阅读:
    matplotlib-形状
    matplotlib-区域填充
    C++文件操作
    画数学公式
    文字
    画注释
    Doubango简介-sip
    boost的asio接收单路大数据量udp包的方法
    Boost.Asio基本原理(CSDN也有Markdown了,好开森)
    boot asio 非阻塞同步编程---非阻塞的accept和receive。
  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9599431.html
Copyright © 2011-2022 走看看