【POJ 1084】 Square Destroyer
【题目链接】
http://poj.org/problem?id=1084
【算法】
迭代加深
【代码】
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std; int i,n,T,step,k,x;
bool dest[]; inline bool is_square(int x,int y,int len)
{
int i,l,r;
bool ret = true;
l = (x - ) * ( * n + ) + y;
r = l + len - ;
for (i = l; i <= r; i++) ret &= (dest[i] ^ );
l = (x + len - ) * ( * n + ) + y;
r = l + len - ;
for (i = l; i <= r; i++) ret &= (dest[i] ^ );
l = n * x + (x - ) * (n + ) + y;
r = n * (x + len - ) + (x + len - ) * (n + ) + y;
for (i = l; i <= r; i += * n + ) ret &= (dest[i] ^ );
l = n * x + (x - ) * (n + ) + y + len;
r = n * (x + len - ) + (x + len - ) * (n + ) + y + len;
for (i = l; i <= r; i += * n + ) ret &= (dest[i] ^ );
return ret;
}
inline bool check()
{
int i,j,k;
for (k = ; k <= n; k++)
{
for (i = ; i <= n - k + ; i++)
{
for (j = ; j <= n - k + ; j++)
{
if (is_square(i,j,k))
return false;
}
}
}
return true;
}
inline bool IDDFS(int dep)
{
int i,j,k,x,y,len,l,r;
if (dep > step)
{
if (check())
return true;
else return false;
}
x = y = len = ;
for (k = ; k <= n; k++)
{
for (i = ; i <= n - k + ; i++)
{
for (j = ; j <= n - k + ; j++)
{
if (is_square(i,j,k))
{
x = i;
y = j;
len = k;
break;
}
}
if (x) break;
}
if (x) break;
}
l = (x - ) * ( * n + ) + y;
r = l + len - ;
for (i = l; i <= r; i++)
{
dest[i] = true;
if (IDDFS(dep+))
return true;
dest[i] = false;
}
l = (x + len - ) * ( * n + ) + y;
r = l + len - ;
for (i = l; i <= r; i++)
{
dest[i] = true;
if (IDDFS(dep+))
return true;
dest[i] = false;
}
l = n * x + (x - ) * (n + ) + y;
r = n * (x + len - ) + (x + len - ) * (n + ) + y;
for (i = l; i <= r; i += * n + )
{
dest[i] = true;
if (IDDFS(dep+))
return true;
dest[i] = false;
}
l = n * x + (x - ) * (n + ) + y + len;
r = n * (x + len - ) + (x + len - ) * (n + ) + y + len;
for (i = l; i <= r; i += * n + )
{
dest[i] = true;
if (IDDFS(dep+))
return true;
dest[i] = false;
}
return false;
} int main()
{ scanf("%d",&T);
while (T--)
{
scanf("%d",&n);
for (i = ; i <= * n * (n + ); i++) dest[i] = false;
scanf("%d",&k);
for (i = ; i <= k; i++)
{
scanf("%d",&x);
dest[x] = true;
}
for (i = ; i <= * n * (n + ); i++)
{
step = i;
if (IDDFS())
break;
}
printf("%d\n",step);
} return ; }
【POJ 1084】 Square Destroyer的更多相关文章
- 【POJ 2942】Knights of the Round Table(双联通分量+染色判奇环)
[POJ 2942]Knights of the Round Table(双联通分量+染色判奇环) Time Limit: 7000MS Memory Limit: 65536K Total Su ...
- 【POJ 2195】 Going Home(KM算法求最小权匹配)
[POJ 2195] Going Home(KM算法求最小权匹配) Going Home Time Limit: 1000MS Memory Limit: 65536K Total Submiss ...
- bzoj 2295: 【POJ Challenge】我爱你啊
2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec Memory Limit: 128 MB Description ftiasch是个十分受女生欢迎的同学,所以 ...
- 【链表】BZOJ 2288: 【POJ Challenge】生日礼物
2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 382 Solved: 111[Submit][S ...
- BZOJ2288: 【POJ Challenge】生日礼物
2288: [POJ Challenge]生日礼物 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 284 Solved: 82[Submit][St ...
- BZOJ2293: 【POJ Challenge】吉他英雄
2293: [POJ Challenge]吉他英雄 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 80 Solved: 59[Submit][Stat ...
- BZOJ2287: 【POJ Challenge】消失之物
2287: [POJ Challenge]消失之物 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 254 Solved: 140[Submit][S ...
- BZOJ2295: 【POJ Challenge】我爱你啊
2295: [POJ Challenge]我爱你啊 Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 126 Solved: 90[Submit][Sta ...
- BZOJ2296: 【POJ Challenge】随机种子
2296: [POJ Challenge]随机种子 Time Limit: 1 Sec Memory Limit: 128 MBSec Special JudgeSubmit: 114 Solv ...
随机推荐
- 华为 荣耀 等手机解锁BootLoader
下载工具按提示操作即可 链接:https://pan.baidu.com/s/1qZezd1q 密码:8pad 备用链接:https://pan.baidu.com/s/1nwv0heD
- Android测试写入文本Log
写入本地SD卡: @SuppressLint("SdCardPath") public void writeFileSdcard(String fileName, String m ...
- 编译Caffe-Win错误集锦
Caffe在Windows下编译还是遇到不少麻烦的... 1.visual studio 2013 error C2371: 'int8_t' : redefinition; 引入的unistd.h文 ...
- theano和keras安装
最近在学深度学习框架,要用到keras库,keras可以搭建在tensorflow和theano上,我电脑装的是Windows,因此决定在电脑上搭建theano框架 下面回顾我的安装过程: 1.安装a ...
- 偏函数应用(Partial Application)和函数柯里化(Currying)
偏函数应用指的是固化函数的一个或一些参数,从而产生一个新的函数.比如我们有一个记录日志的函数: 1: def log(level, message): 2: print level + ": ...
- 常用的 CSS 技巧
1. 黑白图像 这段代码会让你的彩色照片显示为黑白照片,是不是很酷? img.desaturate { filter: grayscale(%); -webkit-filter: grayscale( ...
- 团体程序设计天梯赛-练习集-L1-039. 古风排版
L1-039. 古风排版 中国的古人写文字,是从右向左竖向排版的.本题就请你编写程序,把一段文字按古风排版. 输入格式: 输入在第一行给出一个正整数N(<100),是每一列的字符数.第二行给出一 ...
- C#调用存储过程中事务级临时表返回DataTable列乱序解决办法
string result = strSqlResult.Substring(3).Trim().Replace("\n", "").Replace(" ...
- 谨慎调整内核参数:vm.min_free_kbytes
内核参数:内存相关 内存管理从三个层次管理内存,分别是node, zone ,page; 64位的x86物理机内存从高地址到低地址分为: Normal DMA32 DMA.随着地址降低. [root@ ...
- luogu 2483 K短路 (可持久化左偏树)
题面: 题目大意:给你一张有向图,求1到n的第k短路 $K$短路模板题 假设整个图的边集为$G$ 首先建出以点$n$为根的,沿反向边跑的最短路树,设这些边构成了边集$T$ 那么每个点沿着树边走到点$n ...