nyoj 130 同样的雪花 【哈希】
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同样的雪花
时间限制： 1000 ms | 内存限制： 65535 KB
难度：4
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.题意：雪花有六个角，分别赋给他们长度，依照顺时针输入，问你在输入的雪花中有没有全然一样的.
分析：依照传统的做法时间是O（n^2），由于数据非常大所以说会超时，要换一种方法，要用到散列表（大神们讲的非常具体，我就现丑了）。
这道题的比較也蛮奇特的。
代码1（链表形式）：
#include <cstdio>
#include <cstring>
#define M 20005
using namespace std;
struct node
{
int a[6];
struct node *next;
/* data */
};
node *s[M];
int match(int *temp, int sum){
int i, j;
node *p; p = s[sum]->next;
while(p){
for(i = 0; i < 6; ++ i){
for(j = 0; j < 6; ++ j){
if(temp[j] != p->a[(i+j)%6]) break;
}
if(j == 6) return true;
for(j = 0; j < 6; ++ j){
if(temp[j] != p->a[(i+6-j)%6]) break;
}
if(j == 6) return true;
}
p = p->next;
}
p = new node;
for(i = 0; i < 6; ++ i) p->a[i] = temp[i];
p->next = s[sum]->next;
s[sum]->next = p;
return false;
}
int main(){
int t, n, i, j, temp[6];
scanf("%d", &t);
while(t --){
int sum,flag = 0;
scanf("%d", &n);
for(i = 0; i < M; ++ i){
s[i] = new node; s[i]->next = NULL;
}
while(n --){
sum = 0;
for(i = 0; i < 6; ++i){
scanf("%d", &temp[i]);
sum += temp[i];
}
sum %= M;
if(!flag){
if(match(temp, sum)) flag = 1;
}
}
if(flag) puts("Twin snowflakes found.");
else puts("No two snowflakes are alike.");
}
return 0;
} 代码2（三维数组）：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 20000
int s[M][100][6];
int len[M];
int match(int *a, int *b){
int i, j, k;
for(i = 0; i < 6; i ++){
for(j = 0; j < 6; j ++){
if(a[j] != b[(i+j)%6]) break;
}
if(j == 6) return true;
for(j = 0; j < 6; j ++){
if(a[j] != b[(i+6-j)%6]) break;
}
if(j == 6) return true;
}
return false;
}
int main(){
int t, n, sum, temp[6];
scanf("%d", &t);
while(t --){
int i, j, k, flag = 0;
scanf("%d", &n);
memset(len, 0, sizeof(int)*(M+1));
while(n --){
sum = 0;
for(i = 0; i < 6; i ++){
scanf("%d",&temp[i]);
sum += temp[i];
}
sum %= M;
for (i = 0; i < 6; ++i){
s[sum][len[sum]][i] = temp[i];
}
++len[sum];
}
for(i = 0; i < M; i ++){
if(len[i] >1)
for(j = 0; j < len[i]-1; j ++){
for(k = j+1; k < len[i]; k ++){
if(match(s[i][j], s[i][k])){
flag = 1;
break;
}
}
if(flag) break;
}
if(flag) break;
}
if(flag) puts("Twin snowflakes found.");
else puts("No two snowflakes are alike.");
}
return 0;
}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/118043.html原文链接：https://javaforall.cn