poj 2312 Battle City
题目连接
http://poj.org/problem?id=1840
Battle City
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 4
YBEB
EERE
SSTE
0 0
Sample Output
8
bfs+优先队列。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<set>
using std::set;
using std::sort;
using std::pair;
using std::swap;
using std::multiset;
using std::priority_queue;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 310;
const int INF = 0x3f3f3f3f;
typedef unsigned long long ull;
char G[N][N];
bool vis[N][N];
int H, W;
const int dx[] = { 0, 0, -1, 1 }, dy[] = { -1, 1, 0, 0 };
struct P {
int x, y, s;
P(int i = 0, int j = 0, int k = 0) :x(i), y(j), s(k) {}
inline bool operator<(const P &t) const {
return s > t.s;
}
}S;
int bfs() {
priority_queue<P> q;
q.push(S);
vis[S.x][S.y] = true;
while (!q.empty()) {
P t = q.top(); q.pop();
rep(i, 4) {
int x = t.x + dx[i], y = t.y + dy[i];
if (x < 0 || x >= H || y < 0 || y >= W) continue;
if (vis[x][y] || G[x][y] == 'R' || G[x][y] == 'S') continue;
if (G[x][y] == 'E') {
q.push(P(x, y, t.s + 1));
vis[x][y] = true;
}
if (G[x][y] == 'B') {
vis[x][y] = true;
q.push(P(x, y, t.s + 2));
}
if (G[x][y] == 'T') return t.s + 1;
} }
return -1;
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
while (~scanf("%d %d", &H, &W), H + W) {
rep(i, H) {
scanf("%s", G[i]);
rep(j, W) {
if (G[i][j] == 'Y') S.x = i, S.y = j;
vis[i][j] = false;
}
}
printf("%d\n", bfs());
}
return 0;
}
poj 2312 Battle City的更多相关文章
- poj 2312 Battle City【bfs+优先队列】
Battle City Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7579 Accepted: 2544 Des ...
- POJ - 2312 Battle City BFS+优先队列
Battle City Many of us had played the game "Battle city" in our childhood, and some people ...
- poj 2312 Battle City(优先队列+bfs)
题目链接:http://poj.org/problem?id=2312 题目大意:给出一个n*m的矩阵,其中Y是起点,T是终点,B和E可以走,S和R不可以走,要注意的是走B需要2分钟,走E需要一分钟. ...
- POJ 2312:Battle City(BFS)
Battle City Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9885 Accepted: 3285 Descr ...
- Battle City
Battle City Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7208 Accepted: 2427 Descr ...
- B - Battle City bfs+优先队列
来源poj2312 Many of us had played the game "Battle city" in our childhood, and some people ( ...
- Battle City 优先队列+bfs
Many of us had played the game "Battle city" in our childhood, and some people (like me) e ...
- C - Battle City BFS+优先队列
Many of us had played the game "Battle city" in our childhood, and some people (like me) e ...
- Poj(2312),坦克大战,BFS的变形
题目链接:http://poj.org/problem?id=2312 挺有趣的一道题目,然而很容易WA,我就WA了一次,虽然我Debug的时候已经知道哪里出问题了,就是比如说我搜到B和E时,从B搜第 ...
随机推荐
- 模糊查询(LIKE)and (PATINDEX() . CHARINDEX())
SQL中的模糊查询一般来说使用模糊查询,大家都会想到LIKE select * from table where a like '%字符%' 如果一个SQL语句中用多个 like模糊查询,并且记录条 ...
- 改变ListCtrl某行的背景色或者字体颜色
大家也许熟悉WM_NOTIFY,控件通过WM_NOTIFY向父窗口发送消息.在WM_NOTIFY消息体中,部分控件会发送NM_CUSTOMDRAW告诉父窗口自己需要绘图. 也可以反射NM_CUSTOM ...
- 在C++中调用DLL中的函数 (2)
应用程序使用DLL可以采用两种方式: 一种是隐式链接,另一种是显式链接.在使用DLL之前首先要知道DLL中函数的结构信息. Visual C++6.0在VC\bin目录下提供了一个名为Dumpbin. ...
- Git入门详解
查看分支:git branch 创建分支:git branch <name> 切换分支:git checkout <name> 创建+切换分支:git checkout -b ...
- (笔记)angular 的根据后台StateCode本地显示指定文案
- iOS 8.3 JB ready
Hi, I've been waiting for a very very long time..Now iOS 8.3 is ready. http://www.taig.com/ You guys ...
- windows 2003 修改远程桌面连接数详细方法
Windows Server 2003默认情况下允许远程终端连接的数量是2个用户,我们可以根据需要适当增加远程连接同时在线的用户数. 单击“开始→运行”,输入“gpedit.msc”打开组策略编辑器窗 ...
- 让chrome打开手机网页
在chrome快捷方式上点右键: "C:\Program Files\Google\Chrome\Application\chrome.exe" -user-agent=" ...
- SE11
自省数据及表存储 2014年4月6日 21:37 类似JAVA反射的特性 SAP提供自省数据的机制 这样可以保证程序的灵活性和动态性 1.使用OO中的cl_abap_typed ...
- 基本的Web控件四
基本的Web控件用法二 ListBox控件 页面布局: <div> <h1>ListBox控件</h1> 学生列表: <br/> <asp:Lis ...