NYOJ 284 坦克大战 && POJ 2312 Battle City （广搜+优先队列）

NYOJ 284 坦克大战 && POJ 2312 Battle City （广搜+优先队列）

链接：​​click here~~​​

题意：

描述

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?

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.

输出 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.

样例输入

3 4YBEB EERE SSTE 0 0

样例输出

8

广搜（BFS）+优先队列，从下午4:00多写到现在，用了三种方法，逐渐算是熟悉了这类题的写法，广搜写的有点感觉，继续保持！

【解题思路】广搜+优先队列，枚举所有方向，注意有个耗时的差异，因此很容易想到优先队列 （priority_queue ）,也可以用深搜试一下

代码：

#include #include #include #include #include using namespace std;const int MAX = 305;char map[MAX][MAX];int n,m,u,v,s,e;struct Node{ int x,y,step; friend bool operator < (const Node a,const Node b) { return a.step > b.step; }};int dir[4][3]= {{1,0},{0,1},{-1,0},{0,-1}};bool judg(Node temp){ int x = temp.x; int y = temp.y; if(x<0||x>=n||y<0||y>=m||map[x][y]=='S'||map[x][y]=='R') return false; return true;}Node pre,last;int BFS(){ priority_queue Q; Node temp; Q.push(pre); while(!Q.empty()) { pre=Q.top(); Q.pop(); if((pre.x==last.x)&&(pre.y==last.y)) return pre.step; for(int i = 0; i < 4; i++) { temp.x = pre.x + dir[i][0]; temp.y = pre.y + dir[i][1]; temp. step = pre.step + 1; if(judg(temp)) { if(map[temp.x][temp.y]=='B') temp.step += 1; map[temp.x][temp.y] = 'S'; Q.push(temp); } } } return -1;}void init(){ for(int i = 0; i < n; i++) scanf("%s",map[i]); for(int i = 0; i< n; i++) for(int j = 0; j < m; j++) { if(map[i][j] == 'Y') { pre.x = i; pre.y = j; pre.step=0; } if(map[i][j] == 'T') { last.x = i; last.y = j; last.step=0; } }}int main(){ while(scanf("%d%d",&n,&m),(n||m)) { init(); int dist = BFS(); printf("%d\n",dist); } return 0;}

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