HDUOJ----(1030)Delta-wave

Delta-wave

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4684    Accepted Submission(s): 1777

Problem Description
A triangle field is numbered with successive integers in the way shown on the picture below. 
The traveller needs to go from the cell with number M to the cell with number N. The traveller is able to enter the cell through cell edges only, he can not travel from cell to cell through vertices. The number of edges the traveller passes makes the length of the traveller's route. 
Write the program to determine the length of the shortest route connecting cells with numbers N and M. 
 

 

Input
Input contains two integer numbers M and N in the range from 1 to 1000000000 separated with space(s).
 

 

Output
Output should contain the length of the shortest route.
 

 

Sample Input
6 12
 

 

Sample Output
3
 

 

Source
Ural Collegiate Programming Contest 1998
 
 
 
 
思路：
比如6 16 向右移一格..经过两条边即可...然后将其分层，确定他们的行列来计算相应的值即可。。哎呀，说不清了..
来看代码ba!。。
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 using namespace std;
 7 int sum(int m,int &cc)
 8 {
 9  
10   int rr=(int)sqrt(1.0*m);  //确定她所在第几行..一右边为参数  
11     if(rr*rr!=m)    rr++; 
12      cc=rr*rr-m;
13     if(cc&1)
14       return 2*(rr-1);
15     else 
16         return 2*rr-1;
17 }
18 
19 int main()
20 {
21    int n,m,ncc,mcc,add;
22  while(cin>>m>>n)
23    {
24        int rankn=sum(n,ncc);   //得到n的所在第几行
25        int rankm=sum(m,mcc);   //得到m的所在第几行
26            ncc>>=1;            //得到n所在第几列，以右边为参数
27            mcc>>=1;            //得到m所在第几列，以右边为参数
28    //此处可以进一步优化...现在就不优化了..
29            if(rankn==rankm)
30            cout<<abs(n-m)<<endl;
31        else 
32            if(rankn<rankm)  
33            {
34            
35               if(mcc>=ncc&&mcc<=(rankm-rankn)/2+ncc)
36                       cout<<(rankm-rankn)<<endl;
37               else
38               { 
39                 
40                  if(2*mcc<(rankm-rankn)/2+2*ncc)
41                 {
42                      add=ncc-mcc;
43                      cout<<2*add+(rankm-rankn)<<endl;
44                  }
45                   else
46                   {
47                    add=mcc-((rankm-rankn)/2+ncc);
48                    cout<<2*add+(rankm-rankn)<<endl;
49                   }
50 
51               }
52            }
53         else
54         {
55             //rankn>rankm
56             if(ncc>=mcc&&ncc<=(rankn-rankm)/2+mcc)
57              cout<<(rankn-rankm)<<endl;
58            else
59            {
60                if(2*ncc<(rankn-rankm)/2+2*mcc)
61                { 
62                add=mcc-ncc;
63                cout<<2*add+(rankn-rankm)<<endl;
64                }
65                else
66                {
67                  add=ncc-((rankn-rankm)/2+mcc);
68                  cout<<2*add+rankn-rankm<<endl;
69                }
70            }
71         }
72    /* cout<<0&1<<endl;*/
73 }
74    
75    return 0;
76 }
View Code

 

posted @ 2013-10-27 19:09  龚细军  阅读(270)  评论(0)  编辑  收藏  举报