Battle City BFS+优先队列

Battle City

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

又是一道BFS+优先队列的题，和之前的WAR CHESS类似，但判断条件较少。这次主要是想作个比较，用暴力DFS跑一边，果断TLE。。BFS+优先队列只遍历最少步数之内的点一次，时间效率快很多

dfs length 861 time 1s+

bfs length 1469 time 0.016s memory 1m

#include<stdio.h> #include<string.h>
char a[305][305];
int b[305][305];
int n,m,f,bx,by,min;
void dfs(int x,int y,int s)
{
    if(x<0||y<0||x>=n||y>=m) return;
    if(b[x][y]==1) return;
    if(a[x][y]==‘T‘){
        f=1;
        if(s<min) min=s;
        return;
    }
    else if(a[x][y]==‘S‘){
        b[x][y]=1;
        return;
    }
    else if(a[x][y]==‘R‘){
        b[x][y]=1;
        return;
    }
    else if(a[x][y]==‘B‘){
        s++;
    }
    b[x][y]=1;dfs(x+1,y,s+1);b[x][y]=0;
    b[x][y]=1;dfs(x,y+1,s+1);b[x][y]=0;
    b[x][y]=1;dfs(x-1,y,s+1);b[x][y]=0;
    b[x][y]=1;dfs(x,y-1,s+1);b[x][y]=0;
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)&&!(n==0&&m==0)){
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=0;i<n;i++){
            scanf("%s",a[i]);
            for(j=0;j<m;j++){
                if(a[i][j]==‘Y‘){
                    bx=i;by=j;
                }
            }
        }
        f=0;
        min=200000;
        dfs(bx,by,0);
        if(f==1) printf("%d\n",min);
        else printf("-1\n");
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

char a[305][305];
int b[305][305];
int t[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int n,m,bx,by;

struct Node{
    int x,y,s;
    friend bool operator<(Node a,Node b)
    {
        return a.s>b.s;
    }
}node;

void bfs()
{
    int i,f=0,tx,ty;
    priority_queue<Node> q;
    node.x=bx;
    node.y=by;
    node.s=0;
    b[bx][by]=1;
    q.push(node);
    while(q.size()){
        for(i=0;i<4;i++){
            tx=q.top().x+t[i][0];
            ty=q.top().y+t[i][1];
            if(tx<0||ty<0||tx>=n||ty>=m) continue;
            if(b[tx][ty]==1) continue;
            b[tx][ty]=1;
            if(a[tx][ty]==‘T‘){
                printf("%d\n",q.top().s+1);
                f=1;
                return;
            }
            else if(a[tx][ty]==‘S‘){
                b[tx][ty]=1;
                continue;
            }
            else if(a[tx][ty]==‘R‘){
                b[tx][ty]=1;
                continue;
            }
            else if(a[tx][ty]==‘B‘){
                node.x=tx;
                node.y=ty;
                node.s=q.top().s+2;
                b[tx][ty]=1;
                q.push(node);
            }
            else if(a[tx][ty]==‘E‘){
                node.x=tx;
                node.y=ty;
                node.s=q.top().s+1;
                b[tx][ty]=1;
                q.push(node);
            }
        }
        q.pop();
    }
    if(f==0) printf("-1\n");
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)&&!(n==0&&m==0)){
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=0;i<n;i++){
            scanf("%s",a[i]);
            for(j=0;j<m;j++){
                if(a[i][j]==‘Y‘){
                    bx=i;by=j;
                }
            }
        }
        bfs();
    }
    return 0;
}