【SSLGZ 1344】Knights
来源:互联网 发布:小天才手表没有网络 编辑:程序博客网 时间:2024/06/13 17:04
问题描述
We are given a chess-board of size n*n, from which some fields have been removed. The task is to determine the maximum number of knights that can be placed on the remaining fields of the board in such a way that none of them check each other.
一张大小为n*n的国际象棋棋盘,上面有一些格子被拿走了,棋盘规模n不超过200。马的攻击方向如下图,其中S处为马位置,标有X的点为该马的攻击点。
Fig.1: A knight placed on the field S checks fields marked with x.
Write a program, that:
reads the description of a chess-board with some fields removed, from the input file kni.in,
determines the maximum number of knights that can be placed on the chess-board in such a way that none of them check each other,
writes the result to the output file kni.out.
你的任务是确定在这个棋盘上放置尽可能多的马,并使他们不互相攻击。
输入
The first line of the input file kni.in contains two integers n and m, separated by a single space, 1<=n<=200, 0<=m<n2; n is the chess-board size and m is the number of removed fields. Each of the following m lines contains two integers: x and y, separated by a single space, 1<=x,y<=n – these are the coordinates of the removed fields. The coordinates of the upper left corner of the board are (1,1), and of the bottom right are (n,n). The removed fields are not repeated in the file.
n*n的棋盘,有m个点不能放置骑士
之后m行为不得放置骑士的坐标
输出
The output file kni.out should contain one integer (in the first and only line of the file). It should be the maximum number of knights that can be placed on the given chess-board without checking each other.
样例输入
3 2
1 1
3 3
样例输出
5
算法讨论
这里的方法有点慢,把能放置骑士的点枚举,将它所能攻击的点标号,连接起来,然后用匹配做一遍最大覆盖集,将总点数减去最大覆盖集的点数就是我们求的答案。
#include<cstdio>#include<cstring>using namespace std;int dx[9]={0,-1,-2,-2,-1,1,2,2,1};int dy[9]={0,-2,-1,1,2,2,1,-1,-2};int a[40005][9];bool f[405][405];int num[405][405];bool v[40005];int link[40005];int n,m,s,ans;bool check(int x,int y){ if (x<1||x>n||y<1||y>n) return false; if (f[x][y]) return false; return true;}bool find(int x){ int p; for (int i=1;i<=a[x][0];i++) if (v[a[x][i]]==false) { p=link[a[x][i]]; link[a[x][i]]=x; v[a[x][i]]=true; if ((p==0)||(find(p))) return true; link[a[x][i]]=p; } return false;}int main(){ scanf("%d%d",&n,&m); int x,y; for (int i=1;i<=m;i++) { scanf("%d%d",&x,&y); f[x][y]=true; } s=0; for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) if (f[i][j]==false) { s++; num[i][j]=s; } int p=0; for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) if (f[i][j]==false) { p=num[i][j]; int t=0; for (int k=1;k<=8;k++) { x=i+dx[k]; y=j+dy[k]; if (check(x,y)) { t++; a[p][t]=num[x][y]; } } a[p][0]=t; } for (int i=1;i<=s;i++) { if (a[i][0]==0) continue; memset(v,0,sizeof(v)); if (find(i)) ans++; } ans=ans/2; printf("%d",s-ans); return 0;}
Pixiv ID:46661498
- 【SSLGZ 1344】Knights
- Knights
- Knights
- SSL 1344 Knights 最大匹配
- 【SSLGZ 1618】剑鱼行动
- 【SSLGZ 1615】Frogger
- 【SSLGZ 1763】观光旅游
- 【SSLGZ 1271】 排序I
- 【SSLGZ 1040】合并果子
- 【SSLGZ 1648】丑数
- 【SSLGZ 1715】计算面积
- 【SSLGZ 2058】字符串编辑
- 【SSLGZ 2655】集合问题
- 【SSLGZ 2081】书本整理
- 【SSLGZ 1673】垃圾陷阱
- 【SSLGZ 1612】最优布线问题
- 【SSLGZ 1670】商店选址问题
- 【SSLGZ 1761】城市问题(Floyd)
- jQuery验证控件jquery.validate.js使用说明+中文API
- 监控指定进程
- Zeppelin执行SparkSQL长时间无响应
- 如何停止Handler的消息队列
- 通过base64字符串之间的编码解码实现图片上传
- 【SSLGZ 1344】Knights
- 机器视觉开源代码集合(转)
- java中==和equals的区别以及java中的堆栈
- POJ
- Maven项目导入到eclipse中
- jmeter.bat启动时提示:'findstr' 不是内部或外部命令,也不是可运行的程序或批处理文件。
- 8.6-全栈Java笔记:Math类和枚举
- C++泛型函数模板类型
- 求1+2+3+...+n,要求不能使用乘除法、for、while、if、else、switch、case等关键字及条件判断语句(A?B:C)