NYOJ130 同样的雪花 【Hash】
同样的雪花
- 描写叙述
- You may have heard that no two snowflakes are alike. Your task is to write a program to determine whether this is really true. Your
program will read information about a collection of snowflakes, and search for a pair that may be identical. Each snowflake has six arms. For each snowflake, your program will be provided with a measurement of the length of each of the six arms. Any pair of
snowflakes which have the same lengths of corresponding arms should be flagged by your program as possibly identical.- 输入
- The first line of the input will contain a single interger T(0<T<10),the number of the test cases.
The first line of every test case will contain a single integer n, 0 < n ≤ 100000, the number of snowflakes to follow. This will be followed by n lines, each describing a snowflake. Each snowflake will be described by a line containing six integers (each integer
is at least 0 and less than 10000000), the lengths of the arms of the snow ake. The lengths of the arms will be given in order around the snowflake (either clockwise or counterclockwise), but they may begin with any of the six arms. For example, the same snowflake
could be described as 1 2 3 4 5 6 or 4 3 2 1 6 5. - 输出
- For each test case,if all of the snowflakes are distinct, your program should print the message:
No two snowflakes are alike.
If there is a pair of possibly identical snow akes, your program should print the message:
Twin snowflakes found. - 例子输入
-
1
2
1 2 3 4 5 6
4 3 2 1 6 5 - 例子输出
-
Twin snowflakes found.
- 来源
- POJ
- 上传者
- 张云聪
题意:给定多组包括6组数的序列,推断当中是否有两组是相等的,相等的根据是这两组数正序匹配或者逆序匹配。
题解:以每组的和为key对每片雪花进行哈希,然后从和相等的雪花里进行匹配。
#include <stdio.h>
#include <string.h>
#include <vector> #define maxn 100001
#define MOD 100001 int n;
struct Node {
int a[6];
} snow[maxn];
std::vector<int> f[MOD]; void getData() {
int i, j, sum;
memset(f, 0, sizeof(f));
scanf("%d", &n);
for(i = 0; i < n; ++i) {
for(j = sum = 0; j < 6; ++j) {
scanf("%d", &snow[i].a[j]);
sum += snow[i].a[j];
}
f[sum % MOD].push_back(i);
}
} bool Judge(int x, int y) {
int i, j, count;
for(i = 0; i < 6; ++i) {
if(snow[x].a[i] == snow[y].a[0]) {
for(count = j = 0; j < 6; ++j)
if(snow[x].a[(i+j)%6] == snow[y].a[j])
++count;
if(count == 6) return true; for(count = j = 0; j < 6; ++j)
if(snow[x].a[((i-j)%6+6)%6] == snow[y].a[j])
++count;
if(count == 6) return true;
}
}
return false;
} bool test(int k, int m) {
int i, j;
for(i = 0; i < m; ++i)
for(j = i + 1; j < m; ++j)
if(Judge(f[k][i], f[k][j]))
return true;
return false;
} void solve() {
int i, j, k, ok = 0;
for(i = 0; i < MOD; ++i)
if(!f[i].empty() && test(i, f[i].size())) {
ok = 1; break;
}
printf(ok ? "Twin snowflakes found.\n"
: "No two snowflakes are alike.\n");
} int main() {
// freopen("stdin.txt", "r", stdin);
int t;
scanf("%d", &t);
while(t--) {
getData();
solve();
}
return 0;
}
然后又尝试了下链表:结果TLE
#include <stdio.h>
#include <stdlib.h>
#include <string.h> #define maxn 100001 int n;
struct Node {
int a[6];
Node *next;
};
struct Node2 {
int sum;
Node2 *nextNode2;
Node *next;
} root; void insertNode(Node2 *n2p, Node tmp) {
Node *p = (Node *) malloc(sizeof(Node));
*p = tmp; p->next = n2p->next;
n2p->next = p;
} void insertNode2(int sum, Node tmp) {
Node2 *p = root.nextNode2, *q = &root;
while(true) {
if(!p || p->sum > sum) {
Node2 *temp = (Node2 *) malloc(sizeof(Node2));
temp->sum = sum; temp->nextNode2 = p;
temp->next = NULL; q->nextNode2 = temp;
insertNode(temp, tmp);
return;
} else if(p->sum == sum) {
insertNode(p, tmp);
return;
}
q = p; p = p->nextNode2;
}
} void destoryNode(Node *p) {
Node *q;
while(p) {
q = p; p = p->next;
free(q);
}
} void destoryNode2(Node2 *p) {
Node2 *q;
while(p) {
destoryNode(p->next);
q = p; p = p->nextNode2;
free(q);
}
} void getData() {
int i, j, sum; Node tmp;
destoryNode2(root.nextNode2);
root.nextNode2 = NULL;
root.next = NULL;
root.sum = -1;
scanf("%d", &n);
for(i = 0; i < n; ++i) {
for(j = sum = 0; j < 6; ++j) {
scanf("%d", &tmp.a[j]);
sum += tmp.a[j];
}
insertNode2(sum, tmp);
}
} bool JudgeArr(int a[], int b[]) {
int i, j, cnt;
for(i = 0; i < 6; ++i) {
if(a[i] == b[0]) {
for(j = cnt = 0; j < 6; ++j)
if(a[(i+j)%6] == b[j]) ++cnt;
if(cnt == 6) return true; for(j = cnt = 0; j < 6; ++j)
if(a[(i-j+6)%6] == b[j]) ++cnt;
if(cnt == 6) return true;
}
}
return false;
} bool JudgeList(Node *pn) {
Node *p1, *p2;
for(p1 = pn; p1; p1 = p1->next) {
for(p2 = p1->next; p2; p2 = p2->next)
if(JudgeArr(p1->a, p2->a)) return true;
}
return false;
} void solve() { Node *np;
for(Node2 *n2p = root.nextNode2; n2p; n2p = n2p->nextNode2) {
if(JudgeList(n2p->next)) {
printf("Twin snowflakes found.\n");
return;
}
}
printf("No two snowflakes are alike.\n");
} int main() {
// freopen("stdin.txt", "r", stdin);
int t;
scanf("%d", &t);
while(t--) {
getData();
solve();
}
return 0;
}
NYOJ130 同样的雪花 【Hash】的更多相关文章
- nyoj-130-相同的雪花(hash)
题目链接 /* Name:NYOJ-130-相同的雪花 Copyright: Author: Date: 2018/4/14 15:13:39 Description: 将雪花各个分支上的值加起来,h ...
- 相同的雪花 Hash
相同的雪花 时间限制:1000 ms | 内存限制:65535 KB 难度:4 描述 You may have heard that no two snowflakes are alike. ...
- nyoj130 相同的雪花
相同的雪花 时间限制:1000 ms | 内存限制:65535 KB 难度:4 描述 You may have heard that no two snowflakes are alike. ...
- POJ 3349 HASH
题目链接:http://poj.org/problem?id=3349 题意:你可能听说话世界上没有两片相同的雪花,我们定义一个雪花有6个瓣,如果存在有2个雪花相同[雪花是环形的,所以相同可以是旋转过 ...
- 0x14 hash
被虐爆了 cry 我的hash是真的菜啊... poj3349 肝了一个上午心态崩了...一上午fail了42次我的天,一开始搞了个排序复杂度多了个log,而且是那种可能不同值相等的hash,把12种 ...
- POJ 3349 Snowflake Snow Snowflakes(哈希表)
题意:判断有没有两朵相同的雪花.每朵雪花有六瓣,比较花瓣长度的方法看是否是一样的,如果对应的arms有相同的长度说明是一样的.给出n朵,只要有两朵是一样的就输出有Twin snowflakes fou ...
- 【hash表】收集雪花
[哈希和哈希表]收集雪花 题目描述 不同的雪花往往有不同的形状.在北方的同学想将雪花收集起来,作为礼物送给在南方的同学们.一共有n个时刻,给出每个时刻下落雪花的形状,用不同的整数表示不同的形状.在收集 ...
- Acwing:137. 雪花雪花雪花(Hash表)
有N片雪花,每片雪花由六个角组成,每个角都有长度. 第i片雪花六个角的长度从某个角开始顺时针依次记为ai,1,ai,2,…,ai,6ai,1,ai,2,…,ai,6. 因为雪花的形状是封闭的环形,所以 ...
- POJ 3349:Snowflake Snow Snowflakes(数的Hash)
http://poj.org/problem?id=3349 Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K T ...
随机推荐
- 使用PLSql连接Oracle时报错ORA-12541: TNS: 无监听程序
非常多时候为了优化我们的启动项把oracle的服务禁止了.但是重新启动启动之后使用PLSQL登陆oracle时会出现无监听程序,这说明我们有一些服务没有启动.我们先查看一下oracle的服务是否启动, ...
- HTML页面之间跳转与传值(JS代码)
跳转的方法如下: 方法一: window.location.href = "b.html"; 方法二(返回上一个页面,这个应该不算,先放在这): window.history.ba ...
- Jmail组件发送邮件说明ASP.NET
ASP.Net环境下使用Jmail组件发送邮件2008-01-25 18:59实现过程: 不同于在Asp中使用Jmail,直接使用 Server.CreateObject("Jmail.Me ...
- Silverlight 图表下载到Excel文件中
一.Silverlight xaml.cs文件按钮触发方法 1.//下载图表 private void btnDown_Click(object sender, RoutedEventA ...
- 删除作业计划出错(DELETE语句与 REFERENCE约束"FK_subplan_job_id"冲突。)
删除作业计划出错(DELETE语句与 REFERENCE约束"FK_subplan_job_id"冲突.) use msdb select * from sysmaintplan_plans --查看 ...
- 基于nginx的HLS简单服务器搭建
一,首先搭建nginx服务器: 1.1,选定源码目录 选定目录 /usr/local/HLS cd /usr/local/HLS 1.2,安装PCRE库 cd /usr/local/HLS 到www. ...
- Qt历史版本下载
今天找到一个Qt官方下载任意版本的链接,在这里分享给大家~ http://download.qt.io/archive/qt/ 里面有Qt的历史版本 原文链接http://www.donnyblog. ...
- volatile举列说明const
1.即使本程序中虽然不改变这种类型的值,但别的比如中断程序可能会改变这个值,加上volatile,编译器不优化,每次都重新访问这个值做判断 2.如 unsigned char flag = 1; in ...
- html label 标签的 for 属性
如果您在 label 元素内点击文本,就会触发此控件.就是说,当用户选择该标签时,浏览器就会自动将焦点转到和标签相关的表单控件上. 有两种使用方法: 方法1 使用for属性 <label for ...
- HDU 3001 状压DP
有道状压题用了搜索被队友骂还能不能好好训练了,, hdu 3001 经典的状压dp 大概题意..有n个城市 m个道路 成了一个有向图.n<=10: 然后这个人想去旅行.有个超人开始可以把他扔到 ...