POJ 1383 Labyrinth（迷宫问题，树的最大直径，经典问题）

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最新推荐文章于 2023-12-15 21:24:18 发布

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本文介绍了一种通过两次广度优先搜索(BFS)来确定迷宫中任意两点间最长路径的方法，适用于由点和线构成的迷宫，旨在解决连接特定点所需的最长绳子长度问题。

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Labyrinth（题目链接）

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 4238		Accepted: 1592

Description

The northern part of the Pyramid contains a very large and complicated labyrinth. The labyrinth is divided into square blocks, each of them either filled by rock, or free. There is also a little hook on the floor in the center of every free block. The ACM have found that two of the hooks must be connected by a rope that runs through the hooks in every block on the path between the connected ones. When the rope is fastened, a secret door opens. The problem is that we do not know which hooks to connect. That means also that the neccessary length of the rope is unknown. Your task is to determine the maximum length of the rope we could need for a given labyrinth.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers C and R (3 <= C,R <= 1000) indicating the number of columns and rows. Then exactly R lines follow, each containing C characters. These characters specify the labyrinth. Each of them is either a hash mark (#) or a period (.). Hash marks represent rocks, periods are free blocks. It is possible to walk between neighbouring blocks only, where neighbouring blocks are blocks sharing a common side. We cannot walk diagonally and we cannot step out of the labyrinth.
The labyrinth is designed in such a way that there is exactly one path between any two free blocks. Consequently, if we find the proper hooks to connect, it is easy to find the right path connecting them.

Output

Your program must print exactly one line of output for each test case. The line must contain the sentence "Maximum rope length is X." where Xis the length of the longest path between any two free blocks, measured in blocks.

Sample Input

2
3 3
###
#.#
###
7 6
#######
#.#.###
#.#.###
#.#.#.#
#.....#
#######

Sample Output

Maximum rope length is 0.
Maximum rope length is 8.

Hint

Huge input, scanf is recommended.
If you use recursion, maybe stack overflow. and now C++/c 's stack size is larger than G++/gcc

Source

Central Europe 1999

题意：

问点点连续之间的最大距离是多少。

两个点的距离是 1 ，三个点是 2 ......

思路：

两次进行 BFS 式的搜索，DFS 是递归，数据量大的会炸。。。。。反正题目说了

代码：6440K 907MS

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MYDD=1103;

int n,m,ans;
int firstx,firsty;//记录查找到 '.' 的坐标

char map[MYDD][MYDD];
int vis[MYDD][MYDD];//标记坐标的访问状态和统计距离
int dx[]= {0,0,-1,1};
int dy[]= {-1,1,0,0};
void BFS() {
	queue<int> Q;
	ans=0;
	memset(vis,-1,sizeof(vis));
	vis[firstx][firsty]=0;
	Q.push(firstx);
	Q.push(firsty);
	while(!Q.empty()) {//每次都敲出空！！
		int nowx=Q.front();
		Q.pop();
		int nowy=Q.front();
		Q.pop();
		for(int j=0; j<4; j++) {
			int gx=nowx+dx[j];
			int gy=nowy+dy[j];
			if(map[gx][gy]=='.'&&gx>=0&&gx<m&&gy>=0&&gy<n&&vis[gx][gy]==-1) {
				Q.push(gx);
				Q.push(gy);
				vis[gx][gy]=vis[nowx][nowy]+1;//累加的距离
				if(ans<vis[gx][gy]) {
					firstx=gx;
					firsty=gy;
					ans=vis[gx][gy];
				}
			}
		}
	}
}

int main() {
	int t;
	scanf("%d",&t);
	while(t--) {
		scanf("%d%d",&n,&m);//n列 m行
		for(int k=0; k<m; k++)
			scanf("%s",map[k]);
			
		int flag=1;
		int r=0;
		while(flag) {
			for(int j=0; j<n; j++) {
				if(map[r][j]=='.') {
					firstx=r;
					firsty=j;
					flag=0;
					break;
				}
			}
			r++;
		}
		
		BFS();
		BFS();
		printf("Maximum rope length is %d.\n",ans);
	}
	return 0;
}

后：

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