poj3038 Children of the Candy Corn DFS+BFS

Children of the Candy Corn

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12020 Accepted: 5168

Description

The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit. 

One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.) 

As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.

Input

Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'. 

Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#'). 

You may assume that the maze exit is always reachable from the start point.

Output

For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.

Sample Input

28 8#########......##.####.##.####.##.####.##.####.##...#..##S#E####9 5##########.#.#.#.#S.......E#.#.#.#.##########

Sample Output

37 5 517 17 9

最后一问求最短路直接BFS，前面两个DFS比较好写，不用标记，找到为止，重点在DFS时候的搜索的优先级顺序，靠左墙和靠右墙的时候只是优先级顺序不同，递归的其他

地方都一样，用一个参数pos（pos是地图的绝对方向）将上一步的方向传进去，以绝对方向来讨论表示走的时候的相对上一步的方向

以靠右墙走为例

u: 上 r: 右 d: 下 l: 左

上一步       下一步优先级

u                 r,u,l,d

r                  d,r,u,l

d                 l,d,r,u

l                  u,l,d,r

靠左墙走直接复制粘贴改改就行了

把情况都枚举出来的，代码略长

#include <cstdio>#include <cstring>#include <iostream>#include <queue>using namespace std;int n, m, sx, sy, ex, ey;char mapp[50][50];bool vis[50][50], vismo[50][50];int dis[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};struct node {int x, y;int step;node() {}node(int x, int y, int step) : x(x), y(y), step(step) {}};queue<node> Q;bool check(int x, int y) {if (x < 0 || x >= n || y < 0 || y >= m || vis[x][y])return false;return true;}int cc = 0; bool dfsl(int step, char dir, int x, int y) {if (step > 1600) return  false;if (x == ex && y == ey) {printf("%d ", step);return true;}if (dir == 'u') {/*  不要写成  if (!vis[x][y - 1]) {     if (dfsl(step + 1, 'l', x, y - 1)) {        return true;  }  }  否则发现第一个不行，下一步其他的方向就不会再试了*/if (!vis[x][y - 1] && dfsl(step + 1, 'l', x, y - 1)) { //左 return true;} else if (!vis[x - 1][y] && dfsl(step + 1, 'u', x - 1, y)) {  //上 return true;} else if (!vis[x][y + 1] && dfsl(step + 1, 'r', x, y + 1)) {  //右 return true;} else if (!vis[x + 1][y] && dfsl(step + 1, 'd', x + 1, y)) {  //下 return true;}} else if (dir == 'd') {if (!vis[x][y + 1] && dfsl(step + 1, 'r', x, y + 1)) { //右 return true;} else if (!vis[x + 1][y] && dfsl(step + 1, 'd', x + 1, y)) { //下 return true;} else if (!vis[x][y - 1] && dfsl(step + 1, 'l', x, y - 1)) { //左 return true;} else if (!vis[x - 1][y] && dfsl(step + 1, 'u', x - 1, y)) { //上 return true;}} else if (dir == 'l') {if (!vis[x + 1][y] && dfsl(step + 1, 'd', x + 1, y)) { //下return true;} else if (!vis[x][y - 1] && dfsl(step + 1, 'l', x, y - 1)) { //左 return true;} else if (!vis[x - 1][y] && dfsl(step + 1, 'u', x - 1, y)) {  //上return true;} else if (!vis[x][y + 1] && dfsl(step + 1, 'r', x, y + 1)) {  //右 return true;}} else if (dir == 'r') {if (!vis[x - 1][y] && dfsl(step + 1, 'u', x - 1, y)) {  //上return true;} else if (!vis[x][y + 1] && dfsl(step + 1, 'r', x, y + 1)) { //右return true;} else if (!vis[x + 1][y] && dfsl(step + 1, 'd', x + 1, y)) { //下return true;} else if (!vis[x][y - 1] && dfsl(step + 1, 'l', x, y - 1)) {  //左return true;}}}//copy->pastebool dfsr(int step, char dir, int x, int y) {if (step > 1600) return  false;if (x == ex && y == ey) {printf("%d ", step);return true;}if (dir == 'u') {if (!vis[x][y + 1] && dfsr(step + 1, 'r', x, y + 1)) { //右 return true;} else if (!vis[x - 1][y] && dfsr(step + 1, 'u', x - 1, y)) {  //上 return true;} else if (!vis[x][y - 1] && dfsr(step + 1, 'l', x, y - 1)) { //左 return true;} else if (!vis[x + 1][y] && dfsr(step + 1, 'd', x + 1, y)) {  //下 return true;}} else if (dir == 'd') {if (!vis[x][y - 1] && dfsr(step + 1, 'l', x, y - 1)) { //左 return true;} else if (!vis[x + 1][y] && dfsr(step + 1, 'd', x + 1, y)) { //下 return true;} else if (!vis[x][y + 1] && dfsr(step + 1, 'r', x, y + 1)) { //右 return true;} else if (!vis[x - 1][y] && dfsr(step + 1, 'u', x - 1, y)) {  //上 return true;}} else if (dir == 'l') {if (!vis[x - 1][y] && dfsr(step + 1, 'u', x - 1, y)) {  //上 return true;} else if (!vis[x][y - 1] && dfsr(step + 1, 'l', x, y - 1)) { //左 return true;} else if (!vis[x + 1][y] && dfsr(step + 1, 'd', x + 1, y)) { //下 return true;} else if (!vis[x][y + 1] && dfsr(step + 1, 'r', x, y + 1)) {  //右 return true;}} else if (dir == 'r') {if (!vis[x + 1][y] && dfsr(step + 1, 'd', x + 1, y)) {  //下return true;} else if (!vis[x][y + 1] && dfsr(step + 1, 'r', x, y + 1)) { //右 return true;} else if (!vis[x - 1][y] && dfsr(step + 1, 'u', x - 1, y)) { //上return true;} else if (!vis[x][y - 1] && dfsr(step + 1, 'l', x, y - 1)) {  //左 return true;}}}void bfss() {memcpy(vis, vismo, sizeof(vis));while(!Q.empty()) Q.pop();Q.push(node(sx, sy, 1));vis[sx][sy] = true;while (!Q.empty()) {node tn = Q.front(); Q.pop();for (int i = 0; i < 4; i++) {int nx = tn.x + dis[i][0];int ny = tn.y + dis[i][1];if (nx == ex && ny == ey) {printf("%d\n", tn.step + 1);return ;}if (check(nx, ny)) {vis[nx][ny] = true;Q.push(node(nx, ny, tn.step + 1));}}}}int main(){int T;scanf("%d", &T); //??while (T--) {memset(vismo, false, sizeof(vismo));scanf("%d%d", &m, &n);for (int i = 0; i < n; i++) {scanf("%s", mapp[i]);for (int j = 0; j < m; j++) {if (mapp[i][j] == '#') vismo[i][j] = true;if (mapp[i][j] == 'S') {sx = i; sy = j;}if (mapp[i][j] == 'E'){ex = i; ey = j;}}}//用一个只读的vismo，怕深广搜标记乱了//然而除了广搜并不用标记，用不用都可以memcpy(vis, vismo, sizeof(vis));char pos;int nx, ny;for (int i = 0; i < 4; i++) {nx = sx + dis[i][0];ny = sy + dis[i][1];if (check(nx, ny)) {switch(dis[i][0]) {case 1: pos = 'd'; break;case -1: pos = 'u'; break;default: break;}switch(dis[i][1]) {case 1: pos = 'r'; break;case -1: pos = 'l'; break;default: break;}dfsl(2, pos, nx, ny);break;}}memcpy(vis, vismo, sizeof(vis));dfsr(2, pos, nx, ny);bfss();}return 0;}