PTA A1013

第七天

A1013 Battle Over Cities (25 分)

题目内容

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 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 3
1 2
1 3
1 2 3

Sample Output:

1
0
0

题目分析

不是很难，有关图的问题，除去被破坏的城市后，有多少个连通图就要修多少条路，代码如下。

具体代码

#include<stdio.h>
#include<stdlib.h>
#define MAXSIZE 1005

int N, M, K;
int road[MAXSIZE][MAXSIZE];
int is_visited[MAXSIZE];

void DFS(int n)
{
	is_visited[n] = 1;
	for (int i = 1; i <= N; i++)
	{
		if (road[n][i] != 0 && is_visited[i] == 0)
			DFS(i);
	}
}

int is_all_visit()
{
	for (int i = 1; i <= N; i++)
	{
		if (is_visited[i] == 0)
			return 0;
	}
	return 1;
}

int main(void)
{
	scanf("%d %d %d", &N, &M, &K);
	for (int i = 0; i < M; i++)
	{
		int b, e;
		scanf("%d %d", &b, &e);
		road[b][e] = 1;
		road[e][b] = 1;
	}
	for (int i = 0; i < K; i++)
	{
		for (int i = 1; i <= N; i++)
			is_visited[i] = 0;
		int m;
		scanf("%d", &m);
		is_visited[m] = 1;
		int num = 0;
		while (!is_all_visit())
		{
			num++;
			for (int i = 1; i <= N; i++)
			{
				if (is_visited[i] == 0)
				{
					DFS(i);
					break;
				}
			}
		}
		printf("%d\n", num - 1);
	}
	system("pause");
}

一点想说的话

今天真是太痛苦了，这几天坚持每天两道题，如果能A掉自然很好，但是如果A不掉就觉得很痛苦，每次都感觉自己有思路，但是总是在细节上出问题，看别人的代码也觉得自己和他们的思路是一样的，为什么我的不对呢？？？有的时候能找到bug有的时候找不到真的非常非常痛苦。
其实今天是写了两道题的，但是第二道真的没写出来，别人的代码思路明明和我一样，样例我也过了，可是就是A不了，一开始总是安慰自己C语言刷难度大一点，可是看了一个大佬的C实现代码，愈加觉得自己是纯粹的菜，跟语言没关系，所以心态有点炸。
我觉得我需要调整一下，顺便趁着这几天把C++学完，所以之后几天每天刷一题。