Swust OJ 004 Maze Problem

Time limit(ms): 1000

Memory limit(kb): 65535

Submission: 661

Accepted: 280

Accepted

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Given a maze, find a shortest path from start to goal.

Description

Input consists serveral test cases.

First line of the input contains number of test case T.

For each test case the first line contains two integers N , M ( 1 <= N, M <= 100 ).

Each of the following N lines contain M characters. Each character means a cell of the map.

Here is the definition for chracter.

 

Constraint:

• 
  
• For a character in the map: 
  
  • 
    
  • 'S' : start cell
    
  • 'E' : goal cell
    
  • '-' : empty cell
    
  • '#' : obstacle cell
    
  
  
  
• no two start cell exists.
  
• no two goal cell exists.
  

Input

For each test case print one line containing shortest path. If there exists no path from start to goal, print -1.

Output

1 2 3 4 5 6 7 8

1 5 5 S-### ----- ##--- E#--- ---##

Sample Input

1 2

9

/*
 *  题意：求迷宫入口到出口的最短距离
 *  bfs大水题。。
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <cmath>
#include <queue>
using namespace std;
int dir[4][2] = { {-1,0}, {0,1}, {0,-1}, {1,0} };
char map[101][101];
bool vis[101][101];
int startI, startJ;
int endI, endJ;
int col, row;
struct Node {
	int step;
	int row;
	int col;
};
void bfs() {
	bool flag = true;
	queue<Node>q;
	Node p;
	p.row = startI;
	p.col = startJ;
	p.step = 0;
	for (int i = 0; i < 101; i++)
		for (int j = 0; j < 101; j++)
			vis[i][j] = 0;
	fill(vis, vis + 101 * 101, 0);
	q.push(p);
	vis[p.row][p.col] = 1;
	while (!q.empty()) {
		Node p = q.front();
		q.pop();
		int r = p.row;
		int c = p.col;
		if (r == endI && c == endJ) {
			flag = false;
			cout << p.step << endl;
			break;
		}
		for (int i = 0; i < 4; i++) {
			Node s;
			s.row = r + dir[i][0];
		    s.col = c + dir[i][1];
			if (s.row>=0 && s.col>=0 && s.row<row && s.col<col && !vis[s.row][s.col] && map[s.row][s.col]!='#') {
				vis[s.row][s.col] = 1;
				s.step = p.step + 1;
				q.push(s);
			}
		}
	}
	if (flag) {
		cout << "-1" << endl;
	}
}
int main(){
	int t;

	cin >> t;
	while (t--) {
		cin >> row >> col;
		for (int i = 0; i < row; i++)
			for (int j = 0; j < col; j++) {
				cin >> map[i][j];
				if (map[i][j] == 'S') {
					startI = i;
					startJ = j;
				}
				if (map[i][j] == 'E') {
					endI = i;
					endJ = j;
				}
			}
		bfs();
	}
	return 0;
}