POJ 2312 Battle City（优先队列+bfs）

Problem Description

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

 

题意：

坦克要从起点（Y），到目的地（T），坦克不能通过钢墙（S），河（R），可以在空地在行走（E），射击破坏砖墙（B），射击砖墙时不行走且花费一个单位的时间，在空地上行走时也花费一个单位的时间。求坦克从起点到目的地最少花多少时间，不可达输出－1；

 

思路：

此题用到了广搜的思想，只是在出队时做了处理，利用优先队列让队列中到起点的时间值最小的点先出队。该方法会用到优先队列的STL。

 

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
struct node {
int x,y,s;
friend bool operator <(node a,node b)
{
    return a.s>b.s;
}
};
char map[305][305];
int n,m;
int vis[305][305];
int dir[][2]={
        {0,1},{0,-1},{1,0},{-1,0}
    };
int sx,sy,ex,ey;
bool ok(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m&&(map[x][y]!='S'&&map[x][y]!='R'))
        return true;
    return false ;
}
int bfs()
{
    priority_queue<node> q;
    node head;
    memset(vis,1,sizeof(vis));
    head.x=sx,head.y=sy,head.s=0;
    q.push(head);
    vis[sx][sy]=0;
    while(!q.empty())
    {
     node  f=q.top();
     q.pop();
     if(f.x==ex&&f.y==ey)
        return f.s;
     for(int i=0;i<4;i++)
     {
         int dx=dir[i][0]+f.x,dy=f.y+dir[i][1];
         if(ok(dx,dy)&&vis[dx][dy])
         {
             vis[dx][dy]=0;
              int temp;
              if(map[dx][dy]=='B') temp=2;
              else temp=1;
              node ss;
              ss.x=dx,ss.y=dy,ss.s=f.s+temp;
              q.push(ss);
         }
 
     }
    }
    return -1;
}
int main()
{
while(scanf("%d%d",&n,&m)&&n+m)
{
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            cin>>map[i][j];
            if(map[i][j]=='Y') sx=i,sy=j;
            else if(map[i][j]=='T') ex=i,ey=j;
        }
    }
    printf("%d\n",bfs());
}
 return 0;
}