noip模拟赛 c

分析:一道比较难的爆搜题.首先要把9个块的信息存下来,记录每个块上下左右位置的颜色,然后记录每一排每一列能否操作,之后就是bfs了。在bfs的时候用一个数记录状态,第i位表示原来的第i个块现在在哪个位置,我们可以通过这个状态来解码得到信息,也可以来判重,只是数组开不下,需要用一个map。然后就是如何判断是否连通.首先把所有块还原成一张图,然后我们可以数每个颜色连通的数量,看看是不是等于36就可以了,这里用dfs来判断.
学习了一种比较强的记录状态的方式:记录相对位置,既可以判重,也可以还原图.
#include <cstdio>
#include <map>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; int canx[], cany[], zhuangtai[];
int can[], vis[][], a[][];
map <long long, int> flag; struct node
{
int col[];
}e[]; struct node2
{
long long stu;
int dist;
}; long long hashh(int a, int b, int c, int d, int e, int f, int g, int h, int i)
{
return (((((((a * + b) * + c) * + d) * + e) * + f) * + g) * + h) * + i;
} int dfs(int col, int x, int y, int last)
{
//printf("%d %d %d %d\n", col, x, y, last);
int ret = ;
if (vis[x][y] || (a[x][y] != col && a[x][y] != ) || (a[x][y] == && last == ))
return ret;
vis[x][y] = ;
if (col == a[x][y])
ret++;
if (x > )
ret += dfs(col, x - , y, a[x][y]);
if (x < )
ret += dfs(col, x + , y, a[x][y]);
if (y > )
ret += dfs(col, x, y - , a[x][y]);
if (y < )
ret += dfs(col, x, y + , a[x][y]);
return ret;
} bool check(long long stu)
{
//printf("%lld\n", stu);
int res = ;
memset(can, , sizeof(can));
memset(a, , sizeof(a));
for (int i = ; i >= ; i--)
for (int j = ; j >= ; j--)
{
int x = stu % ;
a[i * ][j * + ] = e[x].col[];
a[i * + ][j * + ] = e[x].col[];
a[i * + ][j * ] = e[x].col[];
a[i * + ][j * + ] = e[x].col[];
stu /= ;
}
for (int i = ; i <= ; i++)
{
for (int j = ; j <= ; j++)
printf("%d ", a[i][j]);
printf("\n");
}
printf("\n");
for (int i = ; i < ; i++)
for (int j = ; j < ; j++)
for (int k = ; k <= ; k++)
if (!can[k] && a[i][j] == k)
{
memset(vis, , sizeof(vis));
res += dfs(k, i, j, );
can[k] = ;
}
if (res == )
return true;
return false;
} void bfs()
{
queue <node2> q;
node2 temp;
temp.stu = ;
temp.dist = ;
q.push(temp);
flag[] = ;
while (!q.empty())
{
node2 u = q.front();
q.pop();
long long stu = u.stu;
int dist = u.dist;
//printf("%lld %d\n", stu, dist);
if (check(stu))
{
printf("%d\n", dist);
return;
}
//printf("flag\n");
for (int i = ; i >= ; i--)
{
zhuangtai[i] = stu % ;
stu /= ;
}
if (!canx[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
//printf("%lld\n", nstu);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
//printf("%lld\n", nstu);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
}
if (!canx[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!canx[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!cany[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!cany[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!cany[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
}
}
} void init()
{
for (int i = ; i < ; i++)
{
int t;
for (int j = ; j < ; j++)
{
char ch;
cin >> ch;
if (ch == 'R')
e[i].col[j] = ;
if (ch == 'G')
e[i].col[j] = ;
if (ch == 'B')
e[i].col[j] = ;
if (ch == 'O')
e[i].col[j] = ;
}
cin >> t;
if (t == )
{
canx[i / ] = ;
cany[i % ] = ;
}
}
} int main()
{
init();
bfs(); return ;
}
noip模拟赛 c的更多相关文章
- NOIP模拟赛20161022
NOIP模拟赛2016-10-22 题目名 东风谷早苗 西行寺幽幽子 琪露诺 上白泽慧音 源文件 robot.cpp/c/pas spring.cpp/c/pas iceroad.cpp/c/pas ...
- contesthunter暑假NOIP模拟赛第一场题解
contesthunter暑假NOIP模拟赛#1题解: 第一题:杯具大派送 水题.枚举A,B的公约数即可. #include <algorithm> #include <cmath& ...
- NOIP模拟赛 by hzwer
2015年10月04日NOIP模拟赛 by hzwer (这是小奇=> 小奇挖矿2(mining) [题目背景] 小奇飞船的钻头开启了无限耐久+精准采集模式!这次它要将原矿运到泛光之源的矿 ...
- 大家AK杯 灰天飞雁NOIP模拟赛题解/数据/标程
数据 http://files.cnblogs.com/htfy/data.zip 简要题解 桌球碰撞 纯模拟,注意一开始就在袋口和v=0的情况.v和坐标可以是小数.为保险起见最好用extended/ ...
- 队爷的讲学计划 CH Round #59 - OrzCC杯NOIP模拟赛day1
题目:http://ch.ezoj.tk/contest/CH%20Round%20%2359%20-%20OrzCC杯NOIP模拟赛day1/队爷的讲学计划 题解:刚开始理解题意理解了好半天,然后发 ...
- 队爷的Au Plan CH Round #59 - OrzCC杯NOIP模拟赛day1
题目:http://ch.ezoj.tk/contest/CH%20Round%20%2359%20-%20OrzCC杯NOIP模拟赛day1/队爷的Au%20Plan 题解:看了题之后觉得肯定是DP ...
- 队爷的新书 CH Round #59 - OrzCC杯NOIP模拟赛day1
题目:http://ch.ezoj.tk/contest/CH%20Round%20%2359%20-%20OrzCC杯NOIP模拟赛day1/队爷的新书 题解:看到这题就想到了 poetize 的封 ...
- CH Round #58 - OrzCC杯noip模拟赛day2
A:颜色问题 题目:http://ch.ezoj.tk/contest/CH%20Round%20%2358%20-%20OrzCC杯noip模拟赛day2/颜色问题 题解:算一下每个仆人到它的目的地 ...
- CH Round #52 - Thinking Bear #1 (NOIP模拟赛)
A.拆地毯 题目:http://www.contesthunter.org/contest/CH%20Round%20%2352%20-%20Thinking%20Bear%20%231%20(NOI ...
- CH Round #49 - Streaming #4 (NOIP模拟赛Day2)
A.二叉树的的根 题目:http://www.contesthunter.org/contest/CH%20Round%20%2349%20-%20Streaming%20%234%20(NOIP 模 ...
随机推荐
- 如何判断js的变量的数据类型
文章首发: http://www.cnblogs.com/sprying/p/4349426.html 本文罗列了一般的Js中类型检测的方法,实际上是每个新手在构建Js知识体系时,都要知晓的,而我只是 ...
- ACM_巧克力
Chocolate,Chocolate Time Limit: 2000/1000ms (Java/Others) Problem Description: 都说发神喜欢吃巧克力,有一次发神徒弟买了一 ...
- 题解报告:hdu 3790 最短路径问题
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3790 Problem Description 给你n个点,m条无向边,每条边都有长度d和花费p,给你起 ...
- Unicode gbk gb2312 编码问题 [转载]
原文地址: http://www.cnblogs.com/csn0721/archive/2013/01/24/2875613.html HTML5 UTF-8 中文乱码 <!DOCTYPE ...
- Spring Cloud (13) 服务网关-路由配置
传统路由配置 所谓传统路由配置方式就是在不依赖于服务发现机制情况下,通过在配置文件中具体制定每个路由表达式与服务实例的映射关系来实现API网关对外部请求的路由.没有Eureka服务治理框架帮助的时候, ...
- VS开发C语言系列(零)-VS2013写C语言错误汇总
错误代码 error C3861:调用函数前未引用 error C4996:调用不安全的函数 error C2668:重载函数不明确 error C3861:"文件名" 找不到标识 ...
- HDU3949 XOR(线性基第k小)
Problem Description XOR is a kind of bit operator, we define that as follow: for two binary base num ...
- mysql中的各种concat
引用:http://www.cnblogs.com/appleat/archive/2012/09/03/2669033.html 一.CONCAT()函数CONCAT()函数用于将多个字符串连接成一 ...
- 关于jquery $.browser 报错问题
在调用 jquery 插件时,出现$.browser 报错,原来是jQuery 从 1.9 版开始,移除了 $.browser 和 $.browser.version 等属性, 取而代之的是 $.su ...
- SQLSERVER SQL性能优化技巧
这篇文章主要介绍了SQLSERVER SQL性能优化技巧,需要的朋友可以参考下 1.选择最有效率的表名顺序(只在基于规则的优化器中有效) SQLSERVER的解析器按照从右到左的顺序处理F ...