The Wall Pushers UVA

考察抽象思维的一道题目，对各个方向进行编号，从w方向开始，将该方向编号为0，逆时针加1作为每个方向的编号，这样就可以保证对于编号i，1<<i也就满足对应方向的对应权值，然后从起始点开始，对于每个方向，判断这个方向是否存在障碍物，如果是存在障碍物的，那么就继续判断是否存在两个连续的障碍物，如果不是，那么就“移动”该障碍物，这个移动反映到数学上，也就是更新各个格子的数值，然后再递归判断，如果没有障碍物，就直接递归判断就行了，最后打印结果，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;int x, y;int area[10][10];const string dir = "WNES";int dx[] = {0,-1,0,1};int dy[] = {-1,0,1,0};int getMinStep(int cur_x,int cur_y){int res = 10;for (int i = 1; i <= 4; i++){if ((area[i][1] & 1) != 1){res = min(res, abs(cur_x - i) + cur_y);}if ((area[i][6] & 4) != 4){res = min(res, abs(cur_x - i) + 7 - cur_y);}}for (int j = 1; j <= 6; j++){if ((area[1][j] & 2) != 2){res = min(res, abs(cur_y - j) + cur_x);}if ((area[4][j] & 8) != 8){res = min(res, abs(cur_y - j) + 5 - cur_x);}}return res;}bool dfs(int step,int cur_x,int cur_y,string s,int up1){if (step == up1&&(cur_x == 0 || cur_x == 5 || cur_y == 0 || cur_y == 7)){cout << s << endl;return true;}int remain = getMinStep(cur_x, cur_y);if (remain + step > up1) return false;for (int i = 0; i < 4; i++){int newx = cur_x + dx[i];int newy = cur_y + dy[i];if (newx >= 0 && newx <= 5 && newy >= 0 && newy <= 7){if ((area[cur_x][cur_y] & (1 << i)) == (1 << i)){if (newx != 0 && newx != 5 && newy != 0 && newy != 7 &&(area[newx][newy] & (1 << i)) != (1 << i)){area[cur_x][cur_y] -= (1 << i);area[newx][newy] = area[newx][newy] - (1 << ((i + 2) % 4)) + (1 << i);area[newx + dx[i]][newy + dy[i]] += (1 << ((i+2)%4));if (dfs(step + 1, newx, newy, s + dir[i],up1)) return true;area[cur_x][cur_y] += (1 << i);area[newx][newy] = area[newx][newy] + (1 << ((i + 2) % 4)) - (1 << i);area[newx + dx[i]][newy + dy[i]] -= (1 << ((i+2)%4));}}else{if (dfs(step + 1, newx, newy, s+dir[i],up1)){return true;}}}}return false;}int main(){while (cin >> y >> x){if (x == 0 && y == 0) break;for (int i = 1; i <= 4; i++){for (int j = 1; j <= 6; j++){cin >> area[i][j];}}for (int up1 = 1;; up1++){if (dfs(0, x, y, "",up1)){break;}}}return 0;}