(NYoj 284)坦克大战--裸BFS ,优先队列
2016-07-30 12:43
387 查看
坦克大战
时间限制:1000 ms | 内存限制:65535 KB
难度:3
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?
输入
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 4
YBEB
EERE
SSTE
0 0
样例输出
8
分析:
裸的BFS,直接使用优先队列即可。进空格需要一步,进普通砖块需要两步。
AC代码:
时间限制:1000 ms | 内存限制:65535 KB
难度:3
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?
输入
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 4
YBEB
EERE
SSTE
0 0
样例输出
8
分析:
裸的BFS,直接使用优先队列即可。进空格需要一步,进普通砖块需要两步。
AC代码:
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; struct node { int x,y,time; friend bool operator < (node a,node b) { return a.time>b.time; } }; char maze[310][310]; int vis [310][310]; int yx,yy; int dx[4]={0,0,-1,1}; int dy[4]={1,-1,0,0}; int m,n; int bfs() { memset(vis,0,sizeof(vis)); node now,next; now.x=yx;now.y=yy;now.time=0; priority_queue<node> pq; while(!pq.empty()) pq.pop(); vis[yx][yy]=1; pq.push(now); while(!pq.empty()) { now=pq.top(); pq.pop(); if(maze[now.x][now.y]=='T'){return now.time;} for(int i=0;i<4;i++) { next.x=now.x+dx[i];next.y=now.y+dy[i]; if(next.x>=0 && next.x<m && next.y>=0 && next.y<n && !vis[next.x][next.y]) { if(maze[next.x][next.y]=='E' || maze[next.x][next.y]=='T'){next.time=now.time+1;vis[next.x][next.y]=1;pq.push(next);} if(maze[next.x][next.y]=='B'){next.time=now.time+2;vis[next.x][next.y]=1;pq.push(next);} } } } return -1; } int main() { while(scanf("%d%d",&m,&n)==2 && n && m) { for(int i=0;i<m;i++) { scanf("%s",maze[i]); for(int j=0;j<n;j++) { if(maze[i][j]=='Y') { yx=i;yy=j; } } } printf("%d\n",bfs()); } return 0; }
相关文章推荐
- PyGobject(四十八)布局容器之HeaderBar
- Java同步锁问题
- 电影<一个都不能少>观后感 ---记码神实训第四天
- 基于itext7导出pdf实现,支持水印、中文等
- GCD
- 第一次系统安装
- tcp客户端和服务端互发消息
- 最长连续上升子序列
- JS实现动画效果框架
- VS2015编译器问题简单记录
- Android Service1
- Android Intent
- 电路组成
- 最长上升子序列
- Linux 下设置防火墙白名单(RHEL 6 和 CentOS 7)
- 1254 - Prison Break
- timestamp 与 rowversion
- python \的作用是继续
- GZIP压缩原理分析(15)——第五章 Deflate算法详解(五06) 预备知识(05) 预备知识总结
- TimeManagerMent Techique