poj 2312 Battle City(bfs+优先级队列)

Battle City

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9200		Accepted: 3058

DescriptionMany 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, youcan 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?InputThe 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' (brickwall), '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.OutputFor 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

tips:坦克大战。S金属，R河流不能走。B要打一下才能走，相当于走到B要花费2.E随便走。

使用广搜求最优解是显然的。但是广搜时，每次搜索的点，都要是同一级别的点。如果普通队列的话有可能前面的点不是最优点。

因此这里用到了优先级队列。建立一个小顶堆

#include<iostream>#include<cstring>#include<queue>using namespace std;int n,m;char mp[333][333];int book[333][333];int sx,sy,ex,ey;struct node{int x,y,step;//因为要建立一个小顶堆friend bool operator <(node n1,node n2){return n1.step>n2.step;}};int go[][2]={-1,0,1,0,0,-1,0,1};int  bfs(int x,int y){priority_queue<node>q;book[x][y]=1;q.push(node{x,y,0});while(!q.empty()){node tt=q.top();q.pop();int x=tt.x;int y=tt.y;if(x==ex&&y==ey){return tt.step;}for(int i=0;i<4;i++){int nx=x+go[i][0];int ny=y+go[i][1];if(nx<1||nx>n||ny<1||ny>m||mp[nx][ny]=='R'||mp[nx][ny]=='S'||book[nx][ny])continue;if(mp[nx][ny]=='B'){book[nx][ny]=1;q.push(node{nx,ny,tt.step+2});}else{book[nx][ny]=1;q.push(node{nx,ny,tt.step+1});}}}return -1;}int main(){while(cin>>n>>m,n+m){memset(book,0,sizeof(book));for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>mp[i][j];if(mp[i][j]=='Y')sx=i,sy=j;if(mp[i][j]=='T')ex=i,ey=j;}}cout<<bfs(sx,sy)<<endl;;}return 0;}