Knight Moves
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 37011 | Accepted: 17105 |
Description
BackgroundMr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him?
The Problem
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.
Input
The input begins with the number n of scenarios on a single line by itself.Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.
Output
For each scenario of the input you have to calculate the minimal amount of knight moves which are necessary to move from the starting point to the ending point. If starting point and ending point are equal,distance is zero. The distance must be written on a single line.Sample Input
3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1Sample Output
5
28
0Source
TUD Programming Contest 2001, Darmstadt, Germany#include<iostream>
#include <queue>
#include<cstring>
using namespace std;
typedef struct Node{
int x,y;
int clas;
}node;
int delt_x[8]={-2,-2,-1,-1,1,1,2,2};
int delt_y[8]={1,-1,2,-2,2,-2,1,-1};
bool map[305][305];
//queue <node> Selected; 这个queue不能在这定义，因为对于每次bfs应该都是一个全新的空的queue，我说怎么跑不起来
int bfs(int x,int y,int tarx,int tary,int L){
if(x==tarx&&y==tary){ return 0;}
queue <node> Selected;
memset(map,false,sizeof(map));
node start,temp,succ;
start.x=x;
start.y=y;
start.clas=0;
Selected.push(start);
while(!Selected.empty()){ //&&Selected.front().x<L&&Selected.front().y<L
temp=Selected.front();
Selected.pop();
//if(temp.x==tarx&&temp.y==tary){ return temp.clas; }
for(int i=0;i<8;i++){
x=temp.x+delt_x[i];
y=temp.y+delt_y[i];
if(x<L&&y<L&&x>=0&&y>=0){//点在棋谱内，合法！才能进入queue 之前怎样都跑不通原因在于没有限制x,y>=0这一条例，导致时间太长
if(map[x][y]==false){
if(x==tarx&&y==tary){ return temp.clas+1;}
else{
map[x][y]=true;
succ.x=x;
succ.y=y;
succ.clas=temp.clas+1;
Selected.push(succ);
}
}
}
}
}
return 0;
}
int main(){
int N;
cin>>N;
int L;
int x,y,tarx,tary;
while(N--){
cin>>L;
cin>>x>>y;
cin>>tarx>>tary;
cout<<bfs(x,y,tarx,tary,L)<<endl;
}
return 0;
}