POJ 2312 Battle City 【BFS+模拟】

Many of us had played the game "Battle city" in our childhood, and some people (like me) even often play it on computer now. 

What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls and brick walls only. Your task is to get a bonus as soon as possible suppose that no enemies will disturb you (See the following picture). 

Your tank can't move through rivers or walls, but it can destroy brick walls by shooting. A brick wall will be turned into empty spaces when you hit it, however, if your shot hit a steel wall, there will be no damage to the wall. In each of your turns, you can choose to move to a neighboring (4 directions, not 8) empty space, or shoot in one of the four directions without a move. The shot will go ahead in that direction, until it go out of the map or hit a wall. If the shot hits a brick wall, the wall will disappear (i.e., in this turn). Well, given the description of a map, the positions of your tank and the target, how many turns will you take at least to arrive there?

Input

The input consists of several test cases. The first line of each test case contains two integers M and N (2 <= M, N <= 300). Each of the following M lines contains N uppercase letters, each of which is one of 'Y' (you), 'T' (target), 'S' (steel wall), 'B' (brick wall), 'R' (river) and 'E' (empty space). Both 'Y' and 'T' appear only once. A test case of M = N = 0 indicates the end of input, and should not be processed.

Output

For each test case, please output the turns you take at least in a separate line. If you can't arrive at the target, output "-1" instead.

Sample Input

3 4
YBEB
EERE
SSTE
0 0

Sample Output

8

思路：这个题类似我做个这个题！https://blog.csdn.net/lookqaq/article/details/81349121，思路一样，遇到S不走，遇到B，走两步，遇到E直接走一步，问最后走的步数。代码如下【板子题】：

//用队列实现BFS 
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
#define MAXN 300+5
int n,m;
char str[MAXN][MAXN];
int vis[MAXN][MAXN]; //标记数组 
struct node{
	int x;
	int y;
	int step;
	bool friend operator < (node a,node b){
		return a.step>b.step;
	}
};
int d[4][2]={-1,0,1,0,0,1,0,-1};
void BFS(int x1,int y1,int x2,int y2 ){
	memset(vis,0,sizeof(vis));
	priority_queue<node>que;
	node e1,e2;
	e1.x=x1,e1.y=y1,e1.step=0;
	que.push(e1);
	vis[x1][y1]=1;
	int ans=-1;
	while(!que.empty()){
		e1=que.top();
		que.pop();
		if(e1.x==x2&&e1.y==y2){
			ans=e1.step;
			break;
		}
		for(int i=0;i<4;i++){
			e2.x=e1.x+d[i][0];
			e2.y=e1.y+d[i][1];
			if(e2.x<0||e2.y<0||e2.x>=n||e2.y>=m||vis[e2.x][e2.y]||str[e2.x][e2.y]=='S'||str[e2.x][e2.y]=='R') continue;//剪枝操作 
			//if(vis[e2.x][e2.y]==1)  continue;
			//if(str[e2.x][e2.y]=='#') continue;
			if(str[e2.x][e2.y]=='B')
			   e2.step=e1.step+2;
			else
			   e2.step=e1.step+1;
			que.push(e2);
			vis[e2.x][e2.y]=1;
		}
	}
	if(ans==-1)
	   cout<<-1<<endl;
	else
	  cout<<ans<<endl;
}
int main(){
	int a1,b,c,d;
	int i,j;
	while(cin>>n>>m&&n&&m){
		for(i=0;i<n;i++)
		   cin>>str[i];
		for(i=0;i<n;i++)  //记录起点和终点 
		  for(j=0;j<m;j++){
			if(str[i][j]=='Y')  a1=i,b=j;
			if(str[i][j]=='T')  c=i,d=j;
		}
	BFS(a1,b,c,d);  //调用BFS 
}
	return 0;
}

ps:你又擦肩而过！