Battle City

这道题目需要优先队列+广搜来求解

至于不能直接用广搜，我想应该是由于其中要经过砖墙时时间消耗为2，这使得广搜中的最优解难以得到

如下图：

颜色的深浅代表加入的先后，第一次加入B E,第二次加入E B，第三次加入E，第四次加入T

那么得到Y B B E T的路径而最优路径是Y E B E T，这就需要使用优先队列

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 4YBEBEERESSTE0 0

Sample Output

8

#include<cstdio>#include<string.h>#include<queue>using namespace std;struct list{int r;int s;int step;friend bool operator < (struct list a,struct list b){return a.step>b.step;}};int main(){int m,n;while(1){scanf("%d%d",&m,&n);if(m==0&&n==0)break;else{int i,j,x,y,tail,top,ans,flag;struct list qu;char an[305][305];bool anmap[305][305];priority_queue <struct list> q;memset(anmap,0,sizeof(anmap));top=0,tail=0,ans=0,flag=0;for(i=0;i<m;i++){getchar();for(j=0;j<n;j++){scanf("%c",&an[i][j]);if(an[i][j]=='Y'){x=i;y=j;} }}qu.r=x;qu.s=y;qu.step=0;q.push(qu);anmap[x][y]=true;tail++;while(top<tail){if(an[q.top().r][q.top().s]=='T'){flag=1;break;} x=q.top().r;y=q.top().s;if(y-1>=0&&an[x][y-1]!='S'&&an[x][y-1]!='R'&&anmap[x][y-1]==false){if(an[x][y-1]=='B')qu.step=q.top().step+2;else if(an[x][y-1]=='E'||an[x][y-1]=='T')qu.step=q.top().step+1;qu.r=x;qu.s=y-1;anmap[x][y-1]=true;q.push(qu);tail++;}if(y+1<n&&an[x][y+1]!='S'&&an[x][y+1]!='R'&&anmap[x][y+1]==false){if(an[x][y+1]=='B')qu.step=q.top().step+2;else if(an[x][y+1]=='E'||an[x][y+1]=='T')qu.step=q.top().step+1;qu.r=x;qu.s=y+1;anmap[x][y+1]=true;q.push(qu);tail++;}if(x-1>=0&&an[x-1][y]!='S'&&an[x-1][y]!='R'&&anmap[x-1][y]==false){if(an[x-1][y]=='B')qu.step=q.top().step+2;else if(an[x-1][y]=='E'||an[x-1][y]=='T')qu.step=q.top().step+1;qu.r=x-1;qu.s=y;anmap[x-1][y]=true;q.push(qu);tail++;}if(x+1<m&&an[x+1][y]!='S'&&an[x+1][y]!='R'&&anmap[x+1][y]==false){if(an[x+1][y]=='B')qu.step=q.top().step+2;else if(an[x+1][y]=='E'||an[x+1][y]=='T')qu.step=q.top().step+1;qu.r=x+1;qu.s=y;anmap[x+1][y]=true;q.push(qu);tail++;}top++;q.pop();}if(flag){printf("%d\n",q.top().step);}elseprintf("-1\n");}}return 0;}