Codeforces Gym 100650D Queens, Knights and Pawns 暴力
Problem D: Queens, Knights and Pawns
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88443#problem/D
Description
You all are familiar with the famous 8-queens problem which asks you to place 8 queens on a chess board so no two attack each other. In this problem, you will be given locations of queens and knights and pawns and asked to find how many of the unoccupied squares on the board are not under attack from either a queen or a knight (or both). We’ll call such squares “safe” squares. Here, pawns will only serve as blockers and have no capturing ability. The board below has 6 safe squares. (The shaded squares are safe.) Q P Q K Recall that a knight moves to any unoccupied square that is on the opposite corner of a 2x3 rectangle from its current position; a queen moves to any square that is visible in any of the eight horizontal, vertical, and diagonal directions from the current position. Note that the movement of a queen can be blocked by another piece, while a knight’s movement can not.
Input
There will be multiple test cases. Each test case will consist of 4 lines. The first line will contain two integers n and m, indicating the dimensions of the board, giving rows and columns, respectively. Neither integer will exceed 1000. The next three lines will each be of the form k r1 c1 r2 c2 · · · rk ck indicating the location of the queens, knights and pawns, respectively. The numbering of the rows and columns will start at one. There will be no more than 100 of any one piece. Values of n = m = 0 indicate end of input.
Output
Each test case should generate one line of the form Board b has s safe squares. where b is the number of the board (starting at one) and you supply the correct value for s.
Sample Input
4 4 2 1 4 2 4 1 1 2 1 2 3 2 3 1 1 2 1 1 1 0 1000 1000 1 3 3 00 0 0
Sample Output
Board 1 has 6 safe squares. Board 2 has 0 safe squares. Board 3 has 996998 safe squares.
HINT
题意
给你一个棋盘,棋盘上面有士兵,有皇后,有马
士兵不会动,皇后攻击范围是八个方向的直线,马是日字
然后问你最后有多少个格子是安全的
题解:
直接暴力就好了
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200051
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** int n,m;
int s[][];
//queens, knights and pawns
int ans=;
void check(int x,int y)
{
if(x<||x>n||y<||y>m)
return;
if(s[x][y]==)
{
s[x][y]=;
ans++;
}
}
void at_queue(int x,int y)
{
check(x,y);
for(int i=x;i<=n;i++)
{
if(s[i][y]==)
break;
check(i,y);
}
for(int i=x;i>=;i--)
{
if(s[i][y]==)
break;
check(i,y);
}
for(int i=y;i>=;i--)
{
if(s[x][i]==)
break;
check(x,i);
}
for(int i=y;i<=m;i++)
{
if(s[x][i]==)
break;
check(x,i);
}
for(int i=x,j=y;i<=n&&j<=m;i++,j++)
{
if(s[i][j]==)
break;
check(i,j);
}
for(int i=x,j=y;i>=&&j>=;j--,i--)
{
if(s[i][j]==)
break;
check(i,j);
}
for(int i=x,j=y;i>=&&j<=m;i--,j++)
{
if(s[i][j]==)
break;
check(i,j);
}
for(int i=x,j=y;i<=n&&j>=;i++,j--)
{
if(s[i][j]==)
break;
check(i,j);
}
}
void at_knight(int x,int y)
{
check(x,y);
check(x+,y+);
check(x+,y+);
check(x-,y+);
check(x-,y+);
check(x+,y-);
check(x+,y-);
check(x-,y-);
check(x-,y-);
}
struct node
{
int x,y;
};
node que[];
node kni[];
int main()
{
int t=;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==&&m==)
break;
ans=;
memset(que,,sizeof(que));
memset(kni,,sizeof(kni));
memset(s,,sizeof(s));
int k1=read();
for(int i=;i<k1;i++)
{
que[i].x=read();
que[i].y=read();
}
int k2=read();
for(int i=;i<k2;i++)
{
kni[i].x=read();
kni[i].y=read();
int x=kni[i].x,y=kni[i].y;
if(s[x][y]==)
{
s[x][y]=;
ans++;
}
}
int k3=read();
for(int i=;i<k3;i++)
{
int x=read();
int y=read();
if(s[x][y]==)
{
s[x][y]=;
ans++;
}
}
for(int i=;i<k1;i++)
at_queue(que[i].x,que[i].y);
for(int i=;i<k2;i++)
at_knight(kni[i].x,kni[i].y);
printf("Board %d has %d safe squares.\n",t++,n*m-ans);
}
}
Codeforces Gym 100650D Queens, Knights and Pawns 暴力的更多相关文章
- sicily 1172. Queens, Knights and Pawns
Description You all are familiar with the famous 8-queens problem which asks you to place 8 queens o ...
- Codeforces Gym 100803F There is No Alternative 暴力Kruskal
There is No Alternative 题目连接: http://codeforces.com/gym/100803/attachments Description ICPC (Isles o ...
- Codeforces Gym 100286J Javanese Cryptoanalysis 傻逼暴力
原题地址:http://codeforces.com/gym/100286/attachments/download/2013/20082009-acmicpc-northeastern-europe ...
- Codeforces Gym 100342C Problem C. Painting Cottages 暴力
Problem C. Painting CottagesTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1 ...
- Codeforces Gym 100513I I. Sale in GameStore 暴力
I. Sale in GameStore Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100513/p ...
- Codeforces Gym 100203G G - Good elements 标记暴力
G - Good elementsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...
- Codeforces Gym 101190M Mole Tunnels - 费用流
题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...
- Codeforces Gym 101252D&&floyd判圈算法学习笔记
一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈 ...
- Codeforces Gym 101623A - 动态规划
题目传送门 传送门 题目大意 给定一个长度为$n$的序列,要求划分成最少的段数,然后将这些段排序使得新序列单调不减. 考虑将相邻的相等的数缩成一个数. 假设没有分成了$n$段,考虑最少能够减少多少划分 ...
随机推荐
- View.VISIBLE、INVISIBLE、GONE的区别
android中UI应用的开发中经常会使用view.setVisibility()来设置控件的可见性,其中该函数有3个可选值,他们有着不同的含义: View.VISIBLE--->可见View. ...
- HDU 1247 Hat’s Words
Hat’s Words Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- HDU 5832 A water problem
A water problem Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)T ...
- JMeter使用jar进行压力测试
最近需要对改造的redis缓存接口做压力测试,使用了开源压力测试工具JMeter,分享一下自己的使用经验,希望能对需要进行压力测试的开发同学有所帮助. JMeter介绍 JMeter是Apache软件 ...
- Reading or Writing to Another Processes Memory in C# z
http://www.jarloo.com/reading-and-writing-to-memory/ Declarations [Flags] public enum ProcessAccessF ...
- Color Length
题意: 给出两个字符串,求把两字符串组成一个字符串使的字符串中的相同字母的最远距离的和最小. 分析: 本题关键在于怎么计算距离和的方法上.dp[i][j]表示处理到长度i的a串,长度j的b串还需要的计 ...
- 【剑指offer 面试题23】从上往下打印二叉树
思路: 没啥好说的,BFS. C++: #include <iostream> #include <queue> using namespace std; struct Tre ...
- qt集成到vs2010
因为要正式开始学习图形处理相关的知识了,所以要搭建开发平台.我所要用到的平台主要是vs和qt.虽然现在已经出了vs2013了,并且外观看起来特别漂亮,但是我所要用到的qt4版本中,没有对于的qtAdd ...
- 【原】Kryo序列化篇
Kryo是一个快速有效的对象图序列化Java库.它的目标是快速.高效.易使用.该项目适用于对象持久化到文件或数据库中或通过网络传输.Kryo还可以自动实现深浅的拷贝/克隆. 就是直接复制一个对象对象到 ...
- ASP.NET MVC中的Json Binding和Validate
引子:电子商务网站支付功能页面往往会有很多信息,对于这些信息的保存,往往是分步完成的,那么使用Ajax最合适不过了,比如其中的收货人信息模块.这些信息的新建和编辑保存都是用Ajax来完成的.那么有几种 ...