POJ 1657 Distance on Chessboard(搜索题)

//搜索题，王用宽搜解决，后用点和点的斜率解决，只有两种情况，要么1，要么2。//车直线判断即可，要么1，要么2。//象斜率判断加所在格子的黑白情况进行判断，我想出了一个好方法。行列同奇同偶，为白色，行列奇偶互异为黑色。//象如果在黑色格子上，他永远到不了白色格子，这是性质。#include<iostream>#include<queue>using namespace std;bool vis[64];int dir[8][2] = {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};//方向数组struct chess{int coord;int step;chess(int Coord,int Step){coord = Coord;step = Step;}};bool legal(int x,int y)//判断点的合法{if(x < 0 || x > 8 || y < 0 || y > 8)return false;return true;}int king(int st,int ed)//宽搜{int x1,y1,x2,y2;memset(vis,0,sizeof(vis));x1 = st / 8;y1 = st % 8;queue<chess> q;q.push(chess(st,0));while(!q.empty()){if(q.front().coord == ed)return q.front().step;int tx = q.front().coord / 8;int ty = q.front().coord % 8;int step = q.front().step;q.pop();for(int i = 0;i < 7;++i){x2 = tx + dir[i][0];y2 = ty + dir[i][1];if(!vis[x2*8+y2] && legal(x2,y2)){vis[x2*8+y2] = 1;q.push(chess(x2*8+y2,step+1));}}}}int queen(int st,int ed){int x1,y1,x2,y2;x1 = st / 8;y1 = st % 8;x2 = ed / 8;y2 = ed % 8;if(x1 == x2 || y1 == y2)return 1;else if(abs(x1-x2) == abs(y1-y2))return 1;else return 2;}int rook(int st,int ed){int x1,y1,x2,y2;x1 = st / 8;y1 = st % 8;x2 = ed / 8;y2 = ed % 8;if(x1 == x2 || y1 == y2)return 1;else return 2;}int bishop(int st,int ed){int x1,y1,x2,y2,color1,color2;x1 = st / 8;y1 = st % 8;x2 = ed / 8;y2 = ed % 8;if(abs(x1-x2) == abs(y1-y2))return 1;if(x1 % 2 == y1 % 2)color1 = 1;else color1 = 0;if(x2 % 2 != y2 % 2)color2 = 0;else color2 = 1;if(color1 != color2)return -1;else return 2;}int main(){//freopen("in.txt","r",stdin);int T;int st,ed;char s1[3],s2[3];scanf("%d",&T);while(T--){scanf("%s%s",s1,s2);st = abs(s1[0] - 'h') + (s1[1] - '0') * 8;ed = abs(s2[0] - 'h') + (s2[1] - '0') * 8;if(st == ed)//唉，忽略了起点和中点一样的情况，贡献1WA{printf("0 0 0 0/n");continue;}printf("%d ",king(st,ed));printf("%d ",queen(st,ed));printf("%d ",rook(st,ed));int B = bishop(st,ed);if(B == -1)printf("Inf/n");else printf("%d/n",B);}return 0;}