| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 20913 | Accepted: 9702 |
Description
Input
Output
Sample Input
3 8 0 0 7 0 100 0 0 30 50 10 1 1 1 1
Sample Output
5 28 0
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#define maxn 350
using namespace std;
int n;
int x1,x2,y1,y2;
struct node{
int x;
int y;
int sum;
node(int a,int b,int c)
{
x=a;
y=b;
sum=c;
}
};
int go[8][2]={{1,2},{1,-2},{2,1},{2,-1},{-1,2},{-1,-2},{-2,1},{-2,-1}};
int vis[310][310];
int BFS()
{
queue<node> que;
memset(vis,0,sizeof vis);
que.push(node(x1,y1,0));
vis[x1][y1]=1;
while(!que.empty())
{
node temp=que.front();
for(int i=0;i<8;i++)
{
int x0=temp.x+go[i][0];
int y0=temp.y+go[i][1];
int sum0=temp.sum+1;
if(x0>=0&&x0<n&&y0>=0&&y0<n)
{
if(x0==x2&&y0==y2) return sum0;
else if(!vis[x0][y0])
{
que.push(node(x0,y0,sum0));
vis[x0][y0]=1;
}
}
}
que.pop();
}
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d%d%d",&n,&x1,&y1,&x2,&y2);
if(x1==x2&&y1==y2)
printf("0\n");
else
printf("%d\n",BFS());
}
return 0;
}
POJ 1915 Knight Moves(BFS+STL),布布扣,bubuko.com
POJ 1915 Knight Moves(BFS+STL)
原文:http://blog.csdn.net/u013986860/article/details/27589499