http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1119

1119: Collecting Coins

Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 144  Solved: 35
[Submit][Status][Web Board]

Description

In a maze of r rows and c columns, your task is to collect as many coins as possible.
Each square is either your start point "S"(which will become empty after you leave), an empty square ".", a coin square "C" (which will become empty after you step on this square and thus collecting the coin), a rock square "O" or an obstacle square "X".
At each step, you can move one square to the up, down, left or right. You cannot leave the maze or enter an obstacle square, but you can push each rock at most once (i.e. You can treat a rock as an obstacle square after you push it).
To push a rock, you must stand next to it. You can only push the rock along the direction you're facing, into an neighboring empty square (you can't push it outside the maze, and you can't push it to a squarecontiaining a coin).For example, if the rock is to your immediate right, you can only push it to its right neighboring square.
Find the maximal number of coins you can collect.
 

Input

The first line of input contains a single integer T (T<=25), the number of test cases. 
Each test case begins with two integers r and c (2<=r,c<=10), then followed by r lines, each with c columns. 
There will be at most 5 rocks and at most 10 coins in each maze.

Output

For each test case, print the maximal number of coins you can collect.

Sample Input

3
3 4
S.OC
..O.
.XCX
4 6
S.X.CC
..XOCC
...O.C
....XC
4 4
.SXC
OO.C
..XX
.CCC

Sample Output

1
6
3

HINT

 

Source

湖南省第八届大学生计算机程序设计竞赛

分析;

BFS

AC代码;

 #include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long using namespace std;
int mp[][];
int vis[][];
int xadd[] = {,-,,};
int yadd[] = {,,,-};
struct node
{
int x;int y;
int is;
node(int _x, int _y, int _is)
{
x = _x;
y = _y;
is = ;
}
};
vector<node> C;
int mx = ;
int ansnum = ;
int n , m ;
void debug()
{
for(int i = ;i <= n;i ++)
{
for(int j = ;j <= m;j ++)
printf("%d ",mp[i][j]);
printf("\n");
}
}
void bfs(int x, int y ,int ans)
{
//printf("%d %d\n",x,y);
if(mx == ansnum)
return;
vector<node> Q;
Q.push_back(node(x,y,));
vis[x][y] = ;
int l = ;
int r = ;
while(l <= r )
{
for(int i = ;i <= ;i ++)
{
int tx = Q[l].x + xadd[i] ;
int ty = Q[l].y + yadd[i] ;
if(mp[tx][ty] >= && !vis[tx][ty])
{
vis[tx][ty] = ;
r ++ ;
if(mp[tx][ty] == )
{
Q.push_back(node(tx,ty,));
ans ++ ;
}
else Q.push_back(node(tx,ty,));
}
}
l ++ ;
}
if(ans > mx)
mx = ans;
for(int i = ;i < C.size();i ++)
{
if(!C[i].is)
{
for(int s = ;s <= ;s ++)
{
int tx = C[i].x + xadd[s];
int ty = C[i].y + yadd[s];
int ttx = C[i].x - xadd[s];
int tty = C[i].y - yadd[s];
//printf("%d %d %d %d\n",tx,ty,ttx,tty);
if(mp[tx][ty] == && vis[ttx][tty] == )
{
mp[tx][ty] = -;
mp[C[i].x][C[i].y] = ;
C[i].is = ;
bfs(C[i].x,C[i].y,ans);
mp[tx][ty] = ;
mp[C[i].x][C[i].y] = ;
C[i].is = ;
}
}
}
}
for(int i = r; i >= ;i --)
{
vis[Q[i].x][Q[i].y] = ;
if(Q[i].is)
{
mp[Q[i].x][Q[i].y] = ;
}
}
}
int main(){
int t ;
scanf("%d",&t);
while(t--)
{
memset(mp,-,sizeof(mp));
memset(vis,,sizeof(vis));
scanf("%d %d",&n,&m);
char str[];
int bex, bey ;
ansnum = ;
C.clear();
for(int i = ;i <= n;i ++)
{
scanf("%s",&str[]);
for(int j = ;j <= m; j ++)
{
if(str[j] == 'S')
{
mp[i][j] = ;
bex = i ;
bey = j ;
}else if(str[j] == 'C')
{
ansnum ++;
mp[i][j] = ;
}else if(str[j] == 'X')
{
mp[i][j] = -;
}else if (str[j] == 'O'){
mp[i][j] = ;
C.push_back(node(i,j,));
}else {
mp[i][j] = ;
}
}
}
mx = ;
bfs(bex,bey,);
printf("%d\n",mx);
} return ;
}

csuoj 1119: Collecting Coins的更多相关文章

  1. CSU 1119 Collecting Coins

    bfs+dfs 很复杂的搜索题. 因为数据很小,rock最多只有5个,coin最多只有10个,移动rock最多4^5=1024种状态: 思路: 每次先把当前状态能拿到的coin拿走,并将地图当前位置设 ...

  2. UVA 12510/CSU 1119 Collecting Coins DFS

    前年的省赛题,难点在于这个石头的推移不太好处理 后来还是看了阳神当年的省赛总结,发现这个石头这里,因为就四五个子,就暴力dfs处理即可.先把石头当做普通障碍,进行一遍全图的dfs或者bfs,找到可以找 ...

  3. Codeforces D. Sorting the Coins

    D. Sorting the Coins time limit per test 1 second memory limit per test 512 megabytes input standard ...

  4. codeforces 876 D. Sorting the Coins

    http://codeforces.com/contest/876/problem/D D. Sorting the Coins time limit per test 1 second memory ...

  5. D. Sorting the Coins

    Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins togeth ...

  6. ACM-ICPC (10/16) Codeforces Round #441 (Div. 2, by Moscow Team Olympiad)

    A. Trip For Meal Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. ...

  7. 湖南省第八届大学生计算机程序设计竞赛(A,B,C,E,F,I,J)

    A 三家人 Description 有三户人家共拥有一座花园,每户人家的太太均需帮忙整理花园.A 太太工作了5 天,B 太太则工作了4 天,才将花园整理完毕.C 太太因为正身怀六甲无法加入她们的行列, ...

  8. Codeforces Round #615 (Div. 3)

    A. Collecting Coins 题目链接:https://codeforces.com/contest/1294/problem/A 题意: 你有三个姐妹她们分别有 a , b , c枚硬币, ...

  9. Codeforces Round#615 Div.3 解题报告

    前置扯淡 真是神了,我半个小时切前三题(虽然还是很菜) 然后就开始看\(D\),不会: 接着看\(E\),\(dp\)看了半天,交了三次还不行 然后看\(F\):一眼\(LCA\)瞎搞,然后\(15m ...

随机推荐

  1. CentOS 6.5 下 CDH 5.2.1 集群安装(一)

    集群节点数量3 个 192.168.1.170 cdh-master 192.168.1.171 cdh-slave-1 192.168.1.171 cdh-slave-2 一.安装CentOS6.5 ...

  2. java基础-泛型2

    浏览以下内容前,请点击并阅读 声明 6 类型推测 java编译器能够检查所有的方法调用和对应的声明来决定类型的实参,即类型推测,类型的推测算法推测满足所有参数的最具体类型,如下例所示: //泛型方法的 ...

  3. Mysql 5.7 使用SSL安全连接

    MySQL默认的数据通道是不加密的,在一些安全性要求特别高的场景下,我们需要配置MySQL端口为SSL,使得数据通道加密处理,避免敏感信息泄漏和被篡改.当然,启用MySQL SSL之后,由于每个数据包 ...

  4. xp系统重绘边框线不显示(首次加载没有触发paint事件)

    同样是,重绘边框事件,win7系统显示正常,而xp系统却不显示,这是什么原因造成的呢? 于是,小编开始百度,不停的查找原因,通过一番查找,小编也意外的收获了一些内容: 例如:窗口的拖动,放大,缩小,等 ...

  5. 工作中的sql语句总结

    1,查找mysql 数据库 自动 添加 序号 字段列1,2,3,4 ) AS rowno,ip,startcount ) b 2,mysql的分页语句 limit后面第一个参数是index,从0开始: ...

  6. [Android]关于filed 遍历资源文件的排序问题

    Field[] svgfields = R.drawable.class.getFields(); listid = new ArrayList<Integer>(); for (Fiel ...

  7. 折腾一两天,终于学会使用grunt压缩合并混淆JS脚本,小激动,特意记录一下+spm一点意外收获

    很长时间没有更新博客了,实在是太忙啦...0.0 ,以下的东西纯粹是记录,不是我原创,放到收藏夹还担心不够,这个以后常用,想来想去,还是放到这里吧,,丢不了..最后一句废话,网上搜集也好原创也罢,能解 ...

  8. Log4J 配置文件全属性详解

    第一步:加入log4j-1.2.8.jar到lib下. 第二步:在CLASSPATH下建立log4j.properties.内容如下: 1 log4j.rootCategory=INFO, stdou ...

  9. PyInstaller打包步骤简记

    pyinstaller 下载地址:http://www.pyinstaller.org/ 下载后用cmd进入解压文件夹 python setup.py install 安装. 最近用pyinstall ...

  10. 记录一次Tomcat内存泄露原因的追溯

    现象:WEB无法访问.SSH无法登陆.桌面登陆验证失败. 重启服务器后登陆正常. cat /var/log/message显示root用户创建了2000多个sessions后显示内存不足. 进入tom ...