Lightoj 1005 Rooks(DP)
A rook is a piece used in the game of chess which is played on a board of square grids. A rook can only move vertically or horizontally from its current position and two rooks attack each other if one is on the path of the other. In the following figure, the dark squares represent the reachable locations for rook R1 from its current position. The figure also shows that the rook R1 and R2 are in attacking positions where R1 and R3 are not. R2 and R3 are also in non-attacking positions.
Now, given two numbers n and k, your job is to determine the number of ways one can put k rooks on an n x n chessboard so that no two of them are in attacking positions.
Input
Input starts with an integer T (≤ 350), denoting the number of test cases.
Each case contains two integers n (1 ≤ n ≤ 30) and k (0 ≤ k ≤ n2).
Output
For each case, print the case number and total number of ways one can put the given number of rooks on a chessboard of the given size so that no two of them are in attacking positions. You may safely assume that this number will be less than 1017.
Sample Input |
Output for Sample Input |
|
8 1 1 2 1 3 1 4 1 4 2 4 3 4 4 4 5 |
Case 1: 1 Case 2: 4 Case 3: 9 Case 4: 16 Case 5: 72 Case 6: 96 Case 7: 24 Case 8: 0 |
题目要求在n*n的棋盘上放k个车,问有多少种方法。
dp[i][j]代表前i行放j个车的方案数。则dp[i][j]=dp[i-1][j]+dp[i-1][j-1]*(n-(j-1));
或者使用组合数学做。答案是C(n,k)*A(n,k)
/* ***********************************************
Author :guanjun
Created Time :2016/6/9 16:02:10
File Name :1005.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
int x,y;
};
struct cmp{
bool operator()(Node a,Node b){
if(a.x==b.x) return a.y> b.y;
return a.x>b.x;
}
}; bool cmp(int a,int b){
return a>b;
}
ll dp[][];
int n,k;
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
int T;
cin>>T;
for(int t=;t<=T;t++){
cin>>n>>k;
printf("Case %d: ",t);
if(k>n){
puts("");continue;
}
cle(dp);
dp[][]=;
for(int i=;i<=n;i++){
for(int j=;j<=i;j++){
if(j)dp[i][j]=dp[i-][j]+dp[i-][j-]*(n-j+);
else dp[i][j]=dp[i-][j];
}
}
printf("%lld\n",dp[n][k]);
}
return ;
}
Lightoj 1005 Rooks(DP)的更多相关文章
- 1005 - Rooks(规律)
1005 - Rooks PDF (English) Statistics Forum Time Limit: 1 second(s) Memory Limit: 32 MB A rook is ...
- LightOJ 1364 树形DP
52张扑克牌,问拿到指定数量的4个花色的最少次数期望是多少,其中拿到joker必须马上将其视作一种花色,且要使后续期望最小. 转移很容易想到,主要是两张joker的处理,一个状态除了普通的4个方向的转 ...
- A Dangerous Maze (II) LightOJ - 1395(概率dp)
A Dangerous Maze (II) LightOJ - 1395(概率dp) 这题是Light Oj 1027的加强版,1027那道是无记忆的. 题意: 有n扇门,每次你可以选择其中一扇.xi ...
- Where to Run LightOJ - 1287(概率dp)
Where to Run LightOJ - 1287(概率dp) 题面长长的,看了半天也没看懂题意 不清楚的地方,如何判断一个点是否是EJ 按照我的理解 在一个EJ点处,要么原地停留五分钟接着走,要 ...
- (light OJ 1005) Rooks dp
http://www.lightoj.com/volume_showproblem.php?problem=1005 PDF (English) Statistics Forum Tim ...
- Light OJ 1005 - Rooks(DP)
题目大意: 给你一个N和K要求确定有多少种放法,使得没有两个车在一条线上. N*N的矩阵, 有K个棋子. 题目分析: 我是用DP来写的,关于子结构的考虑是这样的. 假设第n*n的矩阵放k个棋子那么,这 ...
- Rooks LightOJ - 1005
https://vjudge.net/problem/LightOJ-1005 题意:在n*n的矩形上放k个车,使得它们不能互相攻击,求方案数. ans[i][j]表示在i*i的矩形上放j个车的方案数 ...
- lightoj 1005 组合数学
题目链接:http://lightoj.com/volume_showproblem.php?problem=1005 #include <cstdio> #include <cst ...
- Light oj 1005 - Rooks (找规律)
题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1005 纸上画一下,找了一下规律,Ank*Cnk. //#pragma comm ...
随机推荐
- kafka flumn sparkstreaming java实现监听文件夹内容保存到Phoenix中
ps:具体Kafka Flumn SparkStreaming的使用 参考前几篇博客 2.4.6.4.1 配置启动Kafka (1) 在slave机器上配置broker 1) 点击CDH上的kafk ...
- maven中的groupId和artifactId到底指的是什么?
具体回答如下: groupid和artifactId被统称为“坐标”是为了保证项目唯一性而提出的,如果你要把你项目弄到maven本地仓库去,你想要找到你的项目就必须根据这两个id去查找. groupI ...
- php基础 数组 遍历
//参数默认值// function abc($a,$b,$c=0){// echo $a,$b,$c;// }// abc(1,3); //可变参数//function def(){// $arr= ...
- Python的3种格式化字符串方法
Python中有3种format字符串的方式: 传统C语言式 命名参数 位置参数 1. 传统C语言式 和c语言里面的 sprintf 类似,参数格式也一样 title = "world&qu ...
- JIRA 6.3.6安装
一:下载JIRA 从官网下载:https://www.atlassian.com/software/jira/download 我下载的版本是Linux版的 JIRA 6.3.6 wget http: ...
- 在后台根据单据标识构建单据的DynamicObject,然后调用BOS的保存服务保存单据。
var bussnessInfo = Kingdee.BOS.ServiceHelper.MetaDataServiceHelper.GetFormMetaData(this.Context, &qu ...
- Codevs 3409 搬礼物
时间限制: 1 s 空间限制: 64000 KB 题目等级 : 青铜 Bronze 题目描述 Description 小浣熊松松特别喜欢交朋友,今年松松生日,就有N个朋友给他送礼物.可是要把这些礼 ...
- PatentTips - Reducing Write Amplification in a Flash Memory
BACKGROUND OF THE INVENTION Conventional NAND Flash memories move data in the background to write ov ...
- jackon - com.fasterxml.jackson.databind.exc.InvalidDefinitionException && UnrecognizedPropertyException: Unrecognized field 异常
在用jackson解析json数据是碰到的问题 1.首先是InvalidDefinitionException 测试发现可能是目标类中无无参数构造方法导致异常. 添加无参构造方法后发现前一个异常解决但 ...
- Spring中使用byType实现Beans自动装配
以下内容引用自http://wiki.jikexueyuan.com/project/spring/beans-auto-wiring/spring-autowiring-byType.html: 此 ...