【经典算法问题】马的遍历【回溯】



【【经典算法问题】马的遍历【回溯】】

/* 马的遍历 回溯 在N*M的棋盘中的一点(x,y)开始遍历棋盘所有点 2014-4-8 20:10:48 */ #include #define max 1000struct Node{ int x, y; }moveXY[8] = {{1, 2}, {1, -2}, {-1, 2}, {-1, -2}, {2, -1}, {2, 1}, {-2, 1}, {-2, -1}}; bool hasAccess[max][max]; Node store[max]; int n, m, id; bool checkBound(int x, int y){ if(x < 1 || y < 1 || x > n || y > m) return 0; return 1; }void print(){ for(int i = 1; i <= id; ++i){ if(i % 4 == 0) printf("\n"); printf("(%d, %d) ", store[i].x, store[i].y); } printf("\n\n"); }void backTrack(Node k){ if(id == n * m){ print(); return; } Node temp; for(int i = 0; i < 8; ++i){ if(checkBound(k.x + moveXY[i].x, k.y + moveXY[i].y) && !hasAccess[k.x + moveXY[i].x][k.y + moveXY[i].y]){ hasAccess[k.x + moveXY[i].x][k.y + moveXY[i].y] = 1; temp.x = k.x + moveXY[i].x; temp.y = k.y + moveXY[i].y; store[++id] = temp; backTrack(temp); --id; hasAccess[k.x + moveXY[i].x][k.y + moveXY[i].y] = 0; } } }int main(){ int x, y; scanf("%d%d", &n, &m); scanf("%d%d", &x, &y); Node temp; temp.x = x; temp.y = y; store[++id] = temp; hasAccess[x][y] = 1; backTrack(temp); return 0; }



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