Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.

Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.

Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

Input

The input contains multiple test cases.

Each test case include, first two integers n, m. (2<=n,m<=200).

Next n lines, each line included m character.

‘Y’ express yifenfei initial position.

‘M’    express Merceki initial position.

‘#’ forbid road;

‘.’ Road.

‘@’ KCF

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

Sample Input

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
[email protected]
.#...
.#...
@..M.
#...#

Sample Output

66
88
66

Author

yifenfei

Source

奋斗的年代

简单的BFS题，bfs来记录Y和M到每一个点的最小步数。这里只是要注意可能会有的KFC不能到达

AC代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, m;

int mp[205][205];

int d[205][205];
int vis[205][205];

const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};

struct node {
	int a;
	int b;
	int cnt;
	node (int x, int y, int z) {
		a = x;
		b = y;
		cnt = z;
	}
};

void bfs(int x, int y) {	//bfs遍历来存储M和Y到每一个点的距离
	memset(vis, 0, sizeof(vis));
	queue<node> que;
	que.push(node(x, y, 0));
	vis[x][y] = 1;
	while(!que.empty()) {
		node t = que.front();
		que.pop();
		for(int i = 0; i < 4; i ++) {
			int xx = t.a + dx[i];
			int yy = t.b + dy[i];
			if(xx >= 0 && xx <= n - 1 && yy >= 0 && yy <= m - 1 && mp[xx][yy] != 2 && !vis[xx][yy]) {
				que.push(node(xx, yy, t.cnt + 1));
				d[xx][yy] += t.cnt + 1;
				vis[xx][yy] = 1;
			}
		}
	}
}

int main() {
	while(scanf("%d %d", &n, &m) != EOF) {
		int yx, yy;
		int mx, my;
		char s[205];
		for(int i = 0; i < n; i ++) {
			scanf("%s", &s);
			for(int j = 0; j < m; j ++) {
				if(s[j] == ‘.‘) mp[i][j] = 1;
				else if(s[j] == ‘#‘) mp[i][j] = 2;
				else if(s[j] == ‘@‘) {
					mp[i][j] = 3;
				}
				else if(s[j] == ‘Y‘) {
					mp[i][j] = 4;
					yx = i; yy = j;
				}
				else if(s[j] == ‘M‘) {
					mp[i][j] = 4;
					mx = i; my = j;
				}
			}
		}

//		for(int i = 0; i < n; i ++, cout << endl) {
//			for(int j = 0; j < m; j ++) {
//				cout << mp[i][j] << " ";
//			}
//		}

		memset(d, 0, sizeof(d));
		bfs(yx, yy);
		bfs(mx, my);

		int ans = INF;
		for(int i = 0; i < n; i ++) {
			for(int j = 0; j < m; j ++) {
				if(mp[i][j] == 3) {
					if(d[i][j] != 0) {	//要注意还有可能有的KFC不能到达
						ans = min(ans, d[i][j]);
					}
				}
			}
		}
		printf("%d\n", ans * 11);
	}
	return 0;
}