There are four identical pieces on the board. In one move it is allowed to:
> move a piece to an empty neighboring field (up, down, left or right),
> jump over one neighboring piece to an empty field (up, down, left or right).
There are 4 moves allowed fZ http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vciBlYWNoIHBpZWNlIGluIHRoZSBjb25maWd1cmF0aW9uIHNob3duIGFib3ZlLiBBcyBhbiBleGFtcGxlIGxldA=="s consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right.
Write a program that:
> reads two chessboard configurations from the standard input,
> verifies whether the second one is reachable from the first one in at most 8 moves,
> writes the result to the standard output.
Input Each of two input lines contains 8 integers a1, a2, ..., a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.
Output The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.
Sample Input
4 4 4 5 5 4 6 5 2 4 3 3 3 6 4 6
Sample Output
YES
题意就是有四颗棋子，能否在8步内走到目标状态
开个8维数组记录状态即可
#include#include #include #include using namespace std; bool vis[8][8][8][8][8][8][8][8]; bool map[10][10]; int to[4][2] = {1,0,-1,0,0,1,0,-1}; struct point { int x[4],y[4],step; } s,e; int check(point a)//判断是否为最终态 { for(int i = 0; i<4; i++) { if(!map[a.x[i]][a.y[i]]) return 0; } return 1; } int empty(point a,int k)//看要将要到达的那格是否为空 { for(int i = 0; i<4; i++) { if(i!=k && a.x[i] == a.x[k] && a.y[i] == a.y[k]) return 0; } return 1; } int judge(point next)//判断是否符合要求 { int i; for(i = 0; i<4; i++) { if(next.x[i]<0 || next.x[i]>=8 || next.y[i]<0 || next.y[i]>=8) return 1; } if(vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]][next.x[2]][next.y[2]][next.x[3]][next.y[3]]) return 1; return 0; } int bfs() { memset(vis,false,sizeof(vis)); int i,j; queue Q; point a,next; a.step = 0; for(i = 0; i<4; i++) { a.x[i] = s.x[i]; a.y[i] = s.y[i]; } Q.push(a); vis[a.x[0]][a.y[0]][a.x[1]][a.y[1]][a.x[2]][a.y[2]][a.x[3]][a.y[3]] = true; while(!Q.empty()) { a = Q.front(); Q.pop(); if(a.step>=8)//因为后面循环有判断减枝，所以这里要包括8步 return 0; if(check(a)) return 1; for(i = 0; i<4; i++) { for(j = 0; j<4; j++) { next = a; next.x[i]+=to[j][0]; next.y[i]+=to[j][1]; next.step++; if(judge(next)) continue; if(empty(next,i))//要去的那一格是空 { if(check(next)) return 1; vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]][next.x[2]][next.y[2]][next.x[3]][next.y[3]] = true; Q.push(next); } else//非空则继续往前 { next.x[i]+=to[j][0]; next.y[i]+=to[j][1]; if(judge(next) || !empty(next,i))//继续往前也要满足要求且是空格 continue; if(check(next)) return 1; vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]][next.x[2]][next.y[2]][next.x[3]][next.y[3]] = true; Q.push(next); } } } } return 0; } int main() { int i; while(~scanf("%d%d",&s.x[0],&s.y[0])) { s.x[0]--; s.y[0]--; for(i = 1; i<4; i++) { scanf("%d%d",&s.x[i],&s.y[i]); s.x[i]--; s.y[i]--; } memset(map,false,sizeof(map)); for(i = 0; i<4; i++) { scanf("%d%d",&e.x[i],&e.y[i]); e.x[i]--; e.y[i]--; map[e.x[i]][e.y[i]] = true; } int flag = bfs(); if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }