pku 3083 Children of the Candy Corn DFS+BFS

两次dfs,一次bfs,貌似数据不是很大0ms过.

不过这一题相当于几道题,代码写的挺长...

难点在于dfs时的走向向判断问题, 我采用的方法是:

方向left, up, right, down分别用0, 1, 2, 3表示.

保存上一次的位置,再根据这一次的位置,判断其方向,然后向左旋转(沿着left走时)或向右向右旋转(沿着right走时),找出其下一步前进的方向.

 

#include <iostream>#include <algorithm>#include <vector>#include <queue>using namespace std;int maze[40][40];int w, h;int startx,starty, exitx, exity;int cnt1, cnt2;int dx[] = {-1, 0, 1, 0};int dy[] = {0, -1, 0, 1};inline int judgeDir1(int x,int y, int dir){if(maze[y+dy[(dir-1+4)%4]][x+dx[(dir-1+4)%4]] == 1 || maze[y+dy[(dir-1+4)%4]][x+dx[(dir-1+4)%4]] == 3)return (dir-1+4)%4;else if(maze[y+dy[(dir)%4]][x+dx[(dir)%4]] == 1 ||maze[y+dy[(dir)%4]][x+dx[(dir)%4]] == 3)return (dir)%4;else if(maze[y+dy[(dir+1)%4]][x+dx[(dir+1)%4]] == 1 ||maze[y+dy[(dir+1)%4]][x+dx[(dir+1)%4]] == 3)return (dir+1)%4;elsereturn (dir+2)%4;}inline int judgeDir2(int x,int y, int dir){if(maze[y+dy[(dir+1)%4]][x+dx[(dir+1)%4]] == 1 || maze[y+dy[(dir+1)%4]][x+dx[(dir+1)%4]] == 3)return (dir+1)%4;else if(maze[y+dy[(dir)%4]][x+dx[(dir)%4]] == 1 ||maze[y+dy[(dir)%4]][x+dx[(dir)%4]] == 3)return (dir)%4;else if(maze[y+dy[(dir-1+4)%4]][x+dx[(dir-1+4)%4]] == 1 ||maze[y+dy[(dir-1+4)%4]][x+dx[(dir-1+4)%4]] == 1)return (dir-1+4)%4;elsereturn (dir-2+4)%4;}inline int direction(int x, int y, int ox, int oy){if(y == oy && x < ox)return 0; //leftelse if(x == ox && y < oy)return 1; //upelse if(y == oy && ox < x)return 2; //rightelsereturn 3; //down}void dfs1(int x, int y, int ox, int oy, int n){if(x == exitx && y == exity){cnt1 = n+1;return;}int dir = judgeDir1(x, y, direction(x, y, ox, oy));if(dir == 0)dfs1(x-1, y, x, y, n+1);else if(dir == 1)dfs1(x, y-1, x, y, n+1);else if(dir == 2)dfs1(x+1, y, x, y, n+1);elsedfs1(x, y+1, x, y, n+1);}void dfs2(int x, int y, int ox, int oy, int n){if(x == exitx && y == exity){cnt2 = n+1;return;}int dir = judgeDir2(x, y, direction(x, y, ox, oy));if(dir == 0)dfs2(x-1, y, x, y, n+1);else if(dir == 1)dfs2(x, y-1, x, y, n+1);else if(dir == 2)dfs2(x+1, y, x, y, n+1);elsedfs2(x, y+1, x, y, n+1);}struct Point{int x, y;Point(int xx, int yy){x = xx; y = yy;}};bool ok(int x, int y){if(x < 0 || x >= w || y < 0 || y >= h)return false;return (maze[y][x]==1 || maze[y][x] == 3);}bool visited[40][40];int step[40][40];int bfs(int x, int y){memset(visited, 0, sizeof(visited));memset(step, 0, sizeof(step));queue<Point> q;q.push(Point(x, y));visited[y][x] = true;int cnt = 0;while(!q.empty()){Point p = q.front();if(p.x == exitx && p.y == exity)break;q.pop();for(int i = 0; i < 4; ++i){int tempx = p.x+dx[i], tempy = p.y+dy[i];if(ok(tempx, tempy) && !visited[tempy][tempx]){step[tempy][tempx] = step[p.y][p.x] + 1;q.push(Point(tempx, tempy));visited[tempy][tempx] = true;}}}return step[exity][exitx]+1;}int main(){int cnt;char ch;scanf("%d", &cnt);int orgx, orgy;while(cnt--){scanf("%d%d", &w, &h);for (int i = 0; i < h; ++i)for(int j = 0; j < w; ++j){cin >> ch;if(ch == '#')maze[i][j] = 0;else if(ch == '.')maze[i][j] = 1;else if(ch == 'S'){startx = j, starty = i;maze[i][j] = 2;}else{exitx = j, exity = i;maze[i][j] = 3;}}cnt1 = cnt2 = 0;if(startx == 0)orgx = -1, orgy = starty;else if(startx == w-1)orgx = w, orgy = starty;else if(starty == 0)orgx = startx, orgy = -1;elseorgx = startx, orgy = h;dfs1(startx, starty, orgx, orgy, 0);dfs2(startx, starty, orgx, orgy, 0);printf("%d %d %d/n", cnt1, cnt2, bfs(startx, starty));}return 0;}