Pat(Advanced Level)Practice--1013(Battle Over Cities)

战时高速公路修复算法

最新推荐文章于 2024-02-12 17:49:52 发布

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最新推荐文章于 2024-02-12 17:49:52 发布

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分类专栏： Advance Pat 文章标签： Advance Pat C++ STL BFS DFS

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本文介绍了一种快速确定战时因城市被占领而需修复的高速公路数量的方法。通过使用DFS或BFS算法遍历地图中剩余的高速公路网络，计算出为保持其他城市间连接所需的最少高速公路修复数量。

Pat1013代码

题目描述：

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

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

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

Sample Output

1
0
0

题目其实就是求出联通子集的个数，然后需要几条边将其连接起来，由图论知识知道N个顶点至少需要N-1条边连接成联通图，所以本题DFS或者BFS搜索一遍即可。。。

AC代码：

一：DFS解题

#include<cstdio>
#include<cstring>
#define MAX 1001

using namespace std;

int map[MAX][MAX];
int visited[MAX];

void Init()
{
	int i,j;
	for(i=0;i<MAX;i++)
		for(j=0;j<MAX;j++)
			map[i][j]=0;
}

void DFS(int n,int k)
{
	int i;
	visited[k]=1;
	for(i=1;i<=n;i++)
		if(map[i][k]==1&&!visited[i])
			DFS(n,i);
}

int main(int argc,char *argv[])
{
	int N,M,K;
	int i,j,k;
	int ans;
	Init();
	scanf("%d%d%d",&N,&M,&K);
	for(k=0;k<M;k++)
	{
		scanf("%d%d",&i,&j);
		map[i][j]=1;
		map[j][i]=1;
	}
	while(K--)
	{
		scanf("%d",&k);
		memset(visited,0,sizeof(visited));
		ans=0;
		visited[k]=1;
		for(i=1;i<=N;i++)
		{
			if(!visited[i])
			{
				DFS(N,i);
				ans++;
			}
		}
		printf("%d\n",ans-1);
	}

	return 0;
}

二：BFS解题

#include<cstdio>
#include<cstring>
#include<deque>
#define MAX 1001

using namespace std;

int visited[MAX];
int map[MAX][MAX];

void Init()
{
	int i,j;
	for(i=0;i<MAX;i++)
		for(j=0;j<MAX;j++)
			map[i][j]=0;
}

int main(int argc,char *argv[])
{
	int i,j,k;
	int N,M,K;
	int ans;
	Init();
	scanf("%d%d%d",&N,&M,&K);
	for(k=1;k<=M;k++)
	{
		scanf("%d%d",&i,&j);
		map[i][j]=1;
		map[j][i]=1;
	}
	while(K--)
	{
		scanf("%d",&k);
		memset(visited,0,sizeof(visited));
		visited[k]=1;
		ans=0;
		for(i=1;i<=N;i++)
		{
			if(!visited[i])
			{
				deque<int> q;
				visited[i]=1;
				ans++;
				q.push_back(i);
				while(!q.empty())
				{
					k=q.front();
					q.pop_front();
					for(j=1;j<=N;j++)
					{
						if(!visited[j]&&map[j][k]==1)
						{
							visited[j]=1;
							q.push_back(j);
						}
					}
				}
			}
		}
		printf("%d\n",ans-1);
	}

	return 0;
}