跳马
跳马
时限:1000ms 内存限制:10000K 总时限:3000ms
描述
在国际象棋中,马的走法与中车象棋类似,即俗话说的“马走日”,下图所示即国际象棋中马(K)在一步能到达的格子(其中黑色的格子是能到达的位置)。
![](http://192.168.0.242/images/problemimages/knightmove.bmp)
现有一200*200大小的国际象棋棋盘,棋盘中仅有一个马,给定马的当前位置(S)和目标位置(T),求出马最少需要多少跳才能从当前位置到达目标位置。
![](http://192.168.0.242/images/problemimages/knightmove.bmp)
现有一200*200大小的国际象棋棋盘,棋盘中仅有一个马,给定马的当前位置(S)和目标位置(T),求出马最少需要多少跳才能从当前位置到达目标位置。
输入
本题包含多个测例。输入数据的第一行有一个整数N(1<=N<=1000),表示测例的个数,接下来的每一行有四个以空格分隔的整数,分别表示马当前位置及目标位置的横、纵坐标C(x,y)和G(x,y)。坐标由1开始。
输出
对于每个测例,在单独的一行内输出一个整数,即马从当前位置跳到目标位置最少的跳数。
输入样例
2
1 1 2 1
1 5 5 1
1 1 2 1
1 5 5 1
输出样例
3
4
4
#include <iostream> #include <queue> using namespace std; typedef struct position { int posx; int posy; } position; int main() { int num,startx,starty,endx,endy; cin>>num; while(num>0) { int map[200][200]; cin>>startx>>starty>>endx>>endy; queue<position> minequeue; position startplace; startplace.posx = startx-1; startplace.posy = starty-1; minequeue.push(startplace); for(int i = 0;i < 200; i++) for(int j = 0;j < 200; j++) map[i][j] = -1; map[starty-1][startx-1] = 0; while(!minequeue.empty()) { position temp = minequeue.front(); minequeue.pop(); int posx = temp.posx; int posy = temp.posy; if(posx==endx-1&&posy==endy-1) { cout<<map[posy][posx]<<endl; break; } if(posy-1>=0&&posx-2>=0&&map[posy-1][posx-2]==-1) { position temp1; temp1.posy = posy-1; temp1.posx = posx-2; minequeue.push(temp1); map[posy-1][posx-2] = map[posy][posx]+1; //cout<<map[posy-1][posx-2]<<" "; } if(posy-2>=0&&posx-1>=0&&map[posy-2][posx-1]==-1) { position temp2; temp2.posy = posy-2; temp2.posx = posx-1; minequeue.push(temp2); map[posy-2][posx-1] = map[posy][posx]+1; //cout<<map[posy-2][posx-1]<<" "; } if(posy-2>=0&&posx+1<200&&map[posy-2][posx+1]==-1) { position temp3; temp3.posy = posy-2; temp3.posx = posx+1; minequeue.push(temp3); map[posy-2][posx+1] = map[posy][posx]+1; //cout<<map[posy-2][posx+1]<<" "; } if(posy-1>=0&&posx+2<200&&map[posy-1][posx+2]==-1) { position temp4; temp4.posy = posy-1; temp4.posx = posx+2; minequeue.push(temp4); map[posy-1][posx+2] = map[posy][posx]+1; //cout<<map[posy-1][posx+2]<<" "; } if(posy+2<200&&posx+1<200&&map[posy+2][posx+1]==-1) { position temp5; temp5.posy = posy+2; temp5.posx = posx+1; minequeue.push(temp5); map[posy+2][posx+1] = map[posy][posx]+1; //cout<<map[posy+2][posx+1]<<" "; } if(posy+1<200&&posx+2<200&&map[posy+1][posx+2]==-1) { position temp6; temp6.posy = posy+1; temp6.posx = posx+2; minequeue.push(temp6); map[posy+1][posx+2] = map[posy][posx]+1; //cout<<map[posy+1][posx+2]<<" "; } if(posy+1<200&&posx-2>=0&&map[posy+1][posx-2]==-1) { position temp7; temp7.posy = posy+1; temp7.posx = posx-2; minequeue.push(temp7); map[posy+1][posx-2] = map[posy][posx]+1; //cout<<map[posy+1][posx-2]<<" "; } if(posy+2<200&&posx-1>=0&&map[posy+2][posx-1]==-1) { position temp8; temp8.posy = posy+2; temp8.posx = posx-1; minequeue.push(temp8); map[posy+2][posx-1] = map[posy][posx]+1; //cout<<map[posy+2][posx-1]<<" "; } } num--; } }
态度决定高度,细节决定成败,