题目

解析

标准的BFS题目，在多组输入中要做的就是先找到这一组的起点和终点，然后将其传给bfs，在多组输入中最易忘记的是将数组“归零”【即memset】

代码【结构体】

#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <queue>
#include <climits>  // 包含INT_MAX常量
#include <cctype>
using namespace std;
int t, r, c;
char map[210][210];
int book[210][210];



int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};

//定义结构体用于存储坐标
struct node {
	int x, y, step;
};


//BFS模板
void bfs(int sx, int sy, int ex, int ey) {
	memset(book, 0, sizeof book); //注意：这种多组输入的题目，一定要memset！！！.QAQ
	queue<node> q;
	q.push({sx, sy, 0});
	book[sx][sy] = 1;//标记起点
	while (!q.empty()) {
		node s = q.front();
		q.pop();
		//移动
		for (int i = 0; i < 4; i++) {
			int nx = s.x + dx[i], ny = s.y + dy[i];
			//判定是否满足条件
			if (nx >= 0 && ny >= 0 && nx < r && ny < c && !book[nx][ny] && map[nx][ny] == '.') {
				book[nx][ny] = 1;//标记走过的值
				q.push({nx, ny, s.step + 1}); //传新坐标和步数
			}
			//如果到达目标点，则输出步数
			if (nx == ex && ny == ey) {
				cout << s.step + 1;
				return;
			}
		}
	}
	cout << "oop!";
}

int main() {
	cin >> t;
//多组输入模板
	while (t--) {
		int sx, sy, ex, ey;
		cin >> r >> c;
		for (int i = 0; i < r; i++) {
			for (int j = 0; j < c; j++) {
				cin >> map[i][j];
				if (map[i][j] == 'S') {
					sx = i;
					sy = j;
				} else if (map[i][j] == 'E') {
					ex = i;
					ey = j;
				}
			}
		}
		bfs(sx, sy, ex, ey);
	}
	return 0;
}

用book标记且计步数的代码[非结构体法]

#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <algorithm>
#include <math.h>
#include <queue>
#include <climits>  // 包含INT_MAX常量
#include <cctype>
using namespace std;
int t, r, c;

char map[210][210];
int book[210][210];
typedef pair<int, int> PII;
int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};

void bfs(int sx, int sy, int ex, int ey) {
	memset(book, 0, sizeof book);
	book[sx][sy]=1;
	queue<PII> q;
	q.push({sx, sy});
	while (!q.empty()) {
		PII top = q.front();
		q.pop();
		for (int i = 0; i < 4; i++) {
			int nx = top.first + dx[i], ny = top.second + dy[i];
			if (nx >= 0 && ny >= 0 && nx < r && ny < c && !book[nx][ny] && map[nx][ny] != '#') {
				book[nx][ny] += book[top.first][top.second] + 1;//用book标记且计步数
				q.push({nx, ny});
				if (nx == ex && ny == ey) {
					cout << book[nx][ny] << endl;
					return;
				}

			}
		}
	}
	cout << "oop!" << endl;

}

int main() {
	cin >> t;
	while (t--) {
		cin >> r >> c;
		int sx, sy, ex, ey;
		for (int i = 0; i < r; i++) {
			for (int j = 0; j < c; j++) {
				cin >> map[i][j];
				if (map[i][j] == 'S')
					sx = i, sy = j;
				if (map[i][j] == 'E')
					ex = i, ey = j;
			}
		}
		bfs(sx, sy, ex, ey);
	}
	return 0;
}