The Largest Generation (25)(BFS)(PAT甲级)
#include<bits/stdc++.h>
using namespace std;
int n,m,l,t;
int a[1307][137][67];
int vis[1307][137][67];
typedef struct{
int x,y,z;
}node;
int xx[6]={0,0,0,0,1,-1};
int yy[6]={0,0,1,-1,0,0};
int zz[6]={1,-1,0,0,0,0};
node p;
int bfs(int x,int y,int z){
queue<node>q;
vis[x][y][z]=1;
p.x=x;
p.y=y;
p.z=z;
q.push(p);
int sum=0;
while(!q.empty()){
node tmp=q.front();
q.pop();
sum++;
for(int i=0;i<6;++i){
int tx=tmp.x+xx[i];
int ty=tmp.y+yy[i];
int tz=tmp.z+zz[i];
if(tx>0&&tx<=n&&ty>0&&ty<=m&&tz>0&&tz<=l&&a[tx][ty][tz]&&!vis[tx][ty][tz]){
p.x=tx;
p.y=ty;
p.z=tz;
q.push(p);
vis[tx][ty][tz]=1;
}
}
}
if(sum>=t)
return sum;
return 0;
}
int main(){
std::ios::sync_with_stdio(false);
cin>>n>>m>>l>>t;
for(int i=1;i<=l;++i)
for(int j=1;j<=n;++j)
for(int k=1;k<=m;++k)
cin>>a[j][k][i];
int ans=0;
for(int i=1;i<=l;++i)
for(int j=1;j<=n;++j)
for(int k=1;k<=m;++k)
if(a[j][k][i]&&!vis[j][k][i])
ans+=bfs(j,k,i);
cout<<ans;
return 0;
}
The Largest Generation (25)(BFS)(PAT甲级)的更多相关文章
- PAT Advanced 1094 The Largest Generation (25) [BFS,DFS,树的遍历]
题目 A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level ...
- PTA甲级1094 The Largest Generation (25分)
PTA甲级1094 The Largest Generation (25分) A family hierarchy is usually presented by a pedigree tree wh ...
- pat1094. The Largest Generation (25)
1094. The Largest Generation (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...
- PAT (Advanced Level) Practise - 1094. The Largest Generation (25)
http://www.patest.cn/contests/pat-a-practise/1094 A family hierarchy is usually presented by a pedig ...
- 【PAT甲级】1094 The Largest Generation (25 分)(DFS)
题意: 输入两个正整数N和M(N<100,M<N),表示结点数量和有孩子结点的结点数量,输出拥有结点最多的层的结点数量和层号(根节点为01,层数为1,层号向下递增). AAAAAccept ...
- PAT (Advanced Level) 1094. The Largest Generation (25)
简单DFS. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> ...
- PAT练习——1094 The Largest Generation (25 point(s))
题目如下: #include<iostream> #include<vector> #include<algorithm> using namespace std; ...
- 1094. The Largest Generation (25)
A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level bel ...
- 1006.Sign in and Sign out(25)—PAT 甲级
At the beginning of every day, the first person who signs in the computer room will unlock the door, ...
随机推荐
- 如何设置minSdkVersion和targetSdkVersion
转http://www.07net01.com/2015/07/878098.html minSdkversion和targetSdkVersion相信很多人都不太理解,我在网上也看了许多关于这两者区 ...
- orcal操作锦集
更新时间:update qs_settle_dt_cfg set end_date=to_date('9999-12-31','yyyy-MM-dd');查询时间:select to_char( e ...
- php 微信公众平台开发之微信群发信息
这篇文章主要为大家详细介绍了php微信公众平台开发之微信群发信息,具有一定的参考价值,感兴趣的小伙伴们可以参考一下 1.目的 完成在微信公众号中群发消息.这里只是完成简单的文字发送.也可以发送语音图片 ...
- hbase_异常_02_hbase无法访问16010端口
一.异常现象 上一个异常解决了之后,已经能正常启动hbase了,也能正常使用hbase shell ,但是无法通过浏览器访问 16010端口. 二.异常原因 1.原因一 hbase 1.0 以后的版 ...
- JavaWEB - JSP 指令
- mysql 常用的存储引擎MyISAM/InnoDB比较
- Angular.forEach用法
1.针对对象循环(key,value) 官方示例: var values = {name: 'misko', gender: 'male'}; var log = []; angular.forEac ...
- BZOJ5362: [Lydsy1805月赛]quailty 算法
BZOJ5362: [Lydsy1805月赛]quailty 算法 https://lydsy.com/JudgeOnline/problem.php?id=5362 分析: 题意即求一个最小基环树森 ...
- ACM学习历程—SGU 275 To xor or not to xor(xor高斯消元)
题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=275 这是一道xor高斯消元. 题目大意是给了n个数,然后任取几个数,让他们xor和 ...
- 【LeetCode】012. Integer to Roman
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 t ...