Hdu2612Find a way

题意：

给一个图，Y和M两个人分别出发，想去@，问最短时间能到达哪个@。

思路：2遍bfs

坑点：要判断每个@点两个人是否都可达。

#include <cstdio>
#include <queue> 
#include <cstring>
#include <algorithm>
#define fi first
#define se second
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int maxn = 200+5;
const int INF = 0x3f3f3f3f;

int moved[4][2] = {{0,1}, {1,0}, {0,-1}, {-1,0}};
char G[maxn][maxn];
int n, m, vis[maxn][maxn]; 
int d[2][maxn][maxn];

struct Node{
	int x,y,step;
	Node(int a = 0, int b = 0, int c = 0):x(a),y(b),step(c){}
};
Node Y,M;

void bfs(int cur){
	
	memset(vis, 0, sizeof(vis));
	queue<Node> Q;
	if(cur == 0) Q.push(Y);
	else Q.push(M);
	
	Node t;
	while(!Q.empty()){
		
		t = Q.front(); Q.pop(); 
		for(int i = 0; i < 4; ++i){
			int x2 = t.x + moved[i][0];
			int y2 = t.y + moved[i][1];
			int s2 = t.step + 1;
			if(x2 >= 0&&x2 < n&&y2 >= 0&&y2 < m&&G[x2][y2] != '#'&&!vis[x2][y2]){
				if(G[x2][y2] == '@') d[cur][x2][y2] = s2;
				vis[x2][y2] = 1;
				Q.push(Node(x2,y2,s2));
			} 
		}
	} 
}

int main()
{
	freopen("in.txt","r",stdin);
	
	while(scanf("%d %d",&n,&m) == 2){
		for(int i = 0; i < n; ++i) scanf("%s",G[i]);
		
		for(int i = 0; i < n; ++i){
			for(int j = 0; j < m; ++j){
				if(G[i][j] == 'Y'){
					Y.x = i; Y.y = j; Y.step = 0;
				}
				else if(G[i][j] == 'M'){
					M.x = i; M.y = j; M.step = 0;
				}
			}
		}
		memset(d[0],0,sizeof(d[0])); bfs(0);
		memset(d[1],0,sizeof(d[1])); bfs(1);
		
		int ans = INF;
		for(int i = 0; i < n; ++i){
			for(int j = 0; j < m; ++j){
				if(G[i][j] == '@'&&d[0][i][j] > 0&&d[1][i][j] > 0)
				ans = min(ans, d[0][i][j]+d[1][i][j]);
			}
		}
		
		printf("%d\n",ans*11);
	}

	return 0;
}