博客
关于我
1013 Battle Over Cities
阅读量:429 次
发布时间:2019-03-06

本文共 2276 字,大约阅读时间需要 7 分钟。

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 city​1​​-city​2​​ and city​1​​-city​3​​. Then if city​1​​ is occupied by the enemy, we must have 1 highway repaired, that is the highway city​2​​-city​3​​.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<), 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 K numbers, 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 31 21 31 2 3
 

Sample Output:

100

题意:

几个村庄联通在一起,当敌军占领一个村庄后,这个村庄与外界的联通联通路线被阻断,问要使剩下的村庄联通,需要新修建几条路线。

分析:

使用连接矩阵来表示联通的图,visited[]用来表示当前村庄是否被访问过,利用dfs()来寻找连通分量的个数,连通分量的个数-1就是所需要修建的路的条数。

 

Code:

#include
#include
#include
using namespace std;int v[1010][1010];int visited[1010];int n, k, m, c;void dfs(int node) { visited[node] = 1; for (int i = 1; i <= n; ++i) { if (visited[i] == 0 && v[node][i] == 1) { dfs(i); } }}int main() { scanf("%d%d%d", &n, &k, &m); int x, y; for (int i = 0; i < k; ++i) { scanf("%d%d", &x, &y); v[x][y] = v[y][x] = 1; } for (int i = 0; i < m; ++i) { int cnt = 0; scanf("%d", &c); fill(visited, visited+1010, 0); visited[c] = 1; for (int j = 1; j <= n; ++j) { if (visited[j] == 0) { dfs(j); cnt++; } } cout << cnt-1 << endl; } return 0;}

  

注意:

这里要使用scanf()函数来输入,如果使用cin来输入的话会超时。

 

转载地址:http://dctuz.baihongyu.com/

你可能感兴趣的文章
mysql存储过程批量建表
查看>>
MySQL存储过程的使用实现数据快速插入
查看>>
mysql存储过程详解
查看>>
Mysql存表情符号发生错误
查看>>
MySQL学习-group by和having
查看>>
MySQL学习-MySQL数据库事务
查看>>
MySQL学习-MySQL条件查询
查看>>
MySQL学习-SQL语句的分类与MySQL简单查询
查看>>
MySQL学习-子查询及limit分页
查看>>
MySQL学习-排序与分组函数
查看>>
MySQL学习-连接查询
查看>>
Mysql学习总结(10)——MySql触发器使用讲解
查看>>
Mysql学习总结(11)——MySql存储过程与函数
查看>>
Mysql学习总结(12)——21分钟Mysql入门教程
查看>>
Mysql学习总结(13)——使用JDBC处理MySQL大数据
查看>>
Mysql学习总结(14)——Mysql主从复制配置
查看>>
Mysql学习总结(15)——Mysql错误码大全
查看>>
Mysql学习总结(16)——Mysql之数据库设计规范
查看>>
Mysql学习总结(17)——MySQL数据库表设计优化
查看>>
Mysql学习总结(18)——Mysql主从架构的复制原理及配置详解
查看>>