poj2312-Battle City

Battle City

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9255 Accepted: 3065

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 4YBEBEERESSTE0 0

Sample Output

8

Source

POJ Monthly,鲁小石

思路：因为走一步花费的时间不一定，如果是空格，花费1，如果是砖块， 花费2，所以当前点必须为目前为止花费时间最少的点，即每次寻找最短时间点再BFS。自然而然想到了优先队列。

#include<iostream>#include<queue>#include<cstdio>#include<string>#include<cstring>using namespace std;const int INF=0x3f3f3f3f;const int maxn=305;int n, m, sx, sy, ex, ey;char mp[maxn][maxn];bool vis[maxn][maxn];int dir[4][2]={{0, 1}, {0, -1}, {1, 0}, {-1, 0}};struct node{int x, y, steps;bool operator < (const node &tmp) const {return steps>tmp.steps;}};int BFS(){memset(vis, false, sizeof(vis));priority_queue<node> q;q.push((node){sx, sy, 0});vis[sx][sy]=true;while(q.size()){node tmp;tmp=q.top();q.pop();if(tmp.x==ex && tmp.y==ey){return tmp.steps;}for(int i=0; i<4; i++){int tx=tmp.x+dir[i][0];int ty=tmp.y+dir[i][1];if(tx<0 || tx>=n || ty<0 || ty>=m)continue;if((mp[tx][ty]=='E'||mp[tx][ty]=='T') && vis[tx][ty]==false){q.push((node){tx, ty, tmp.steps+1});vis[tx][ty]=true;}else if(mp[tx][ty]=='B' && vis[tx][ty]==false){q.push((node){tx, ty, tmp.steps+2});vis[tx][ty]=true;}} } return -1; }int main(){while(~scanf("%d%d", &n, &m) && (n||m)){for(int i=0; i<n; i++){scanf("%s", mp+i);for(int j=0; j<m; j++){if(mp[i][j]=='Y'){sx=i; sy=j;}else if(mp[i][j]=='T'){ex=i; ey=j; }}}printf("%d\n", BFS());}return 0;}