hdu1030
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Delta-wave
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5858 Accepted Submission(s): 2219
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.
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
源代码:
#include<iostream>
using namespace std;
using namespace std;
int c1,c2,h1,h2,x;
void load(int c,int h,int s)
{
if(h==h2){
if(c-c2>0)
x=s+c-c2;
else x=s+c2-c;
return ;
}
if(c%2==1)
load(c+1,h+1,s+1);
else if(c2-(h2-h)>c)
load(c+1,h,s+1);
else
load(c-1,h,s+1);
}
void search(int num,int c)
{
int i,j,k;
for(k=0,j=i=1;;i++,j+=2){
k+=j;
if(num<=k&&c==1){
h1=i;
c1=num-(k-j);
break;
}
else if(num<=k&&c==2){
h2=i;
c2=num-(k-j);
break;
}
}
}
int main()
{
int m,n,temp;
while(1)
{
cin>>m>>n;
if(m>n) {temp=m;m=n;n=temp;}
search(m,1);
search(n,2);
load(c1,h1,0);
cout<<x<<endl;
}
return 0;
}
void load(int c,int h,int s)
{
if(h==h2){
if(c-c2>0)
x=s+c-c2;
else x=s+c2-c;
return ;
}
if(c%2==1)
load(c+1,h+1,s+1);
else if(c2-(h2-h)>c)
load(c+1,h,s+1);
else
load(c-1,h,s+1);
}
void search(int num,int c)
{
int i,j,k;
for(k=0,j=i=1;;i++,j+=2){
k+=j;
if(num<=k&&c==1){
h1=i;
c1=num-(k-j);
break;
}
else if(num<=k&&c==2){
h2=i;
c2=num-(k-j);
break;
}
}
}
int main()
{
int m,n,temp;
while(1)
{
cin>>m>>n;
if(m>n) {temp=m;m=n;n=temp;}
search(m,1);
search(n,2);
load(c1,h1,0);
cout<<x<<endl;
}
return 0;
}
这道题难点在于衔接点为三角形,不能直接像下棋的程序一样直接用上下左右递归,需要对数字所在行的位置的奇偶进行判别。挺水的一道题
0 0
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