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            poj3349

            Snowflake Snow Snowflakes

            Time Limit: 4000MS Memory Limit: 65536K
            Total Submissions: 22730 Accepted: 5923

            Description

            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.

            Input

            The first line of input 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.

            Output

            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.

            Sample Input

            2
            1 2 3 4 5 6
            4 3 2 1 6 5

            Sample Output

            Twin snowflakes found.
             
            忽然發現自己已經好長時間沒寫過hash的題目了
            這題寫了半天,覺著寫的特別混亂,然后刪掉重新寫
            這樣看起來簡單多了
            用的直接相加求和模大質數的hash
            #include <cstdio>
            #include 
            <cstdlib>
            #include 
            <cstring>
            #include 
            <cmath>
            #include 
            <ctime>
            #include 
            <cassert>
            #include 
            <iostream>
            #include 
            <sstream>
            #include 
            <fstream>
            #include 
            <map>
            #include 
            <set>
            #include 
            <vector>
            #include 
            <queue>
            #include 
            <algorithm>
            #include 
            <iomanip>
            #define maxn 100001
            #define bp 999983
            using namespace std;
            struct node
            {
                
            int x[6];
                
            int next;
            }
             s[maxn],tmp;
            int sec;
            struct node2
            {
                
            int x[6];
                
            void init()
                
            {
                    
            for(int i=0; i<6; i++) scanf("%d",&x[i]);
                }

            }
             a[maxn];
            int n;
            int head[1000001];
            bool flag[1000001];
            int hs;
            inline 
            int hash(node2 t)
            {
                
            return (t.x[0]%bp+t.x[1]%bp+t.x[2]%bp+t.x[3]%bp+t.x[4]%bp+t.x[5]%bp)%bp;
            }

            void add(int hhh,node t)
            {
                s[sec]
            =t;
                s[sec].next
            =head[hhh];
                head[hhh]
            =sec++;
            }

            bool pans(node t1,node t2)
            {
                
            if(t1.x[0]==t2.x[0]&&t1.x[1]==t2.x[1]&&t1.x[2]==t2.x[2]&&t1.x[3]==t2.x[3]&&t1.x[4]==t2.x[4]&&t1.x[5]==t2.x[5]) return 1;
                
            if(t1.x[0]==t2.x[1]&&t1.x[1]==t2.x[2]&&t1.x[2]==t2.x[3]&&t1.x[3]==t2.x[4]&&t1.x[4]==t2.x[5]&&t1.x[5]==t2.x[0]) return 1;
                
            if(t1.x[0]==t2.x[2]&&t1.x[1]==t2.x[3]&&t1.x[2]==t2.x[4]&&t1.x[3]==t2.x[5]&&t1.x[4]==t2.x[0]&&t1.x[5]==t2.x[1]) return 1;
                
            if(t1.x[0]==t2.x[3]&&t1.x[1]==t2.x[4]&&t1.x[2]==t2.x[5]&&t1.x[3]==t2.x[0]&&t1.x[4]==t2.x[1]&&t1.x[5]==t2.x[2]) return 1;
                
            if(t1.x[0]==t2.x[4]&&t1.x[1]==t2.x[5]&&t1.x[2]==t2.x[0]&&t1.x[3]==t2.x[1]&&t1.x[4]==t2.x[2]&&t1.x[5]==t2.x[3]) return 1;
                
            if(t1.x[0]==t2.x[5]&&t1.x[1]==t2.x[0]&&t1.x[2]==t2.x[1]&&t1.x[3]==t2.x[2]&&t1.x[4]==t2.x[3]&&t1.x[5]==t2.x[4]) return 1;
                
            return 0;
            }

            bool pann(node t1,node t2)
            {
                node tmp1;
                
            for(int i=0;i<6;i++) tmp1.x[i]=t1.x[5-i];    
                
            if(tmp1.x[0]==t2.x[0]&&tmp1.x[1]==t2.x[1]&&tmp1.x[2]==t2.x[2]&&tmp1.x[3]==t2.x[3]&&tmp1.x[4]==t2.x[4]&&tmp1.x[5]==t2.x[5]) return 1;
                
            if(tmp1.x[0]==t2.x[1]&&tmp1.x[1]==t2.x[2]&&tmp1.x[2]==t2.x[3]&&tmp1.x[3]==t2.x[4]&&tmp1.x[4]==t2.x[5]&&tmp1.x[5]==t2.x[0]) return 1;
                
            if(tmp1.x[0]==t2.x[2]&&tmp1.x[1]==t2.x[3]&&tmp1.x[2]==t2.x[4]&&tmp1.x[3]==t2.x[5]&&tmp1.x[4]==t2.x[0]&&tmp1.x[5]==t2.x[1]) return 1;
                
            if(tmp1.x[0]==t2.x[3]&&tmp1.x[1]==t2.x[4]&&tmp1.x[2]==t2.x[5]&&tmp1.x[3]==t2.x[0]&&tmp1.x[4]==t2.x[1]&&tmp1.x[5]==t2.x[2]) return 1;
                
            if(tmp1.x[0]==t2.x[4]&&tmp1.x[1]==t2.x[5]&&tmp1.x[2]==t2.x[0]&&tmp1.x[3]==t2.x[1]&&tmp1.x[4]==t2.x[2]&&tmp1.x[5]==t2.x[3]) return 1;
                
            if(tmp1.x[0]==t2.x[5]&&tmp1.x[1]==t2.x[0]&&tmp1.x[2]==t2.x[1]&&tmp1.x[3]==t2.x[2]&&tmp1.x[4]==t2.x[3]&&tmp1.x[5]==t2.x[4]) return 1;
                
            return 0;
            }

            int main()
            {
                scanf(
            "%d",&n);
                
            for(int i=1; i<=n; i++) a[i].init();
                memset(flag,
            0,sizeof(flag));
                memset(head,
            -1,sizeof(head));
                sec
            =0;
                
            bool isex;
                isex
            =false;
                
            for(int i=1; i<=n; i++)
                
            {
                    hs
            =hash(a[i]);
                    
            if(!flag[hs])
                    
            {
                        flag[hs]
            =1;
                        
            for(int j=0; j<6; j++) tmp.x[j]=a[i].x[j];
                        add(hs,tmp);
                    }

                    
            else
                    
            {
                        
            for(int j=0; j<6; j++) tmp.x[j]=a[i].x[j];
                        
            for(int j=head[hs]; j!=-1; j=s[j].next)
                        
            {
                            
            if(pans(tmp,s[j])||pann(tmp,s[j]))
                            
            {
                                isex
            =1;
                                printf(
            "Twin snowflakes found.\n");
                                
            return 0;
                            }

                            
            else 
                            
            {
                                add(hs,tmp);
                            }

                        }

                    }

                }

                
            if(isex==0) printf("No two snowflakes are alike.\n");
                
            return 0;
            }

            posted on 2012-07-24 13:30 jh818012 閱讀(255) 評論(0)  編輯 收藏 引用

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            • 1.?re: poj1426
            • 我嚓,,輝哥,,居然搜到你的題解了
            • --season
            • 2.?re: poj3083
            • @王私江
              (8+i)&3 相當于是 取余3的意思 因為 3 的 二進制是 000011 和(8+i)
            • --游客
            • 3.?re: poj3414[未登錄]
            • @王私江
              0ms
            • --jh818012
            • 4.?re: poj3414
            • 200+行,跑了多少ms呢?我的130+行哦,你菜啦,哈哈。
            • --王私江
            • 5.?re: poj1426
            • 評論內容較長,點擊標題查看
            • --王私江
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