POJ 3083 Children of the Candy Corn
来源：互联网 发布：印度教 基督教 知乎 编辑：程序博客网 时间：2024/06/02 17:41
一个模拟+bfs就可以水过但是代码量稍微有点大啊、、、不多说了啊满满的都是泪啊，由于bfs标记的错误，RE了十次啊！（bfs每次入队都要进行标记啊！！一定要记住啊！！！）后来又是跳出循环break的位置写错了啊，又是WA啊，写了前前后后八九个小时啊！不停地出错啊、、、菜鸟伤不起啊，智商真是着急啊、、、
Children of the Candy Corn
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8108 Accepted: 3542
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.
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.
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
#include <stdio.h>#include <string.h>#define MAX 110#define MAXN 1100000struct node{ int x1, y1;} bfs[MAXN];int num[MAXN];int map[MAX][MAX], dp[MAX][MAX], f[MAX][MAX];char s[MAX][MAX];int main(){ int i, j, t, tt, n, m, k, x, y, a, b, p, q; int lcnt, rcnt, cnt; scanf("%d",&k); while(k--) { memset(dp , -1 , sizeof(dp)); memset(f , -1 , sizeof(dp)); memset(map , -1 , sizeof(map)); lcnt = 1; rcnt = 1; cnt = 0; scanf("%d %d",&m, &n); for(i = 1; i <= n; i++) scanf("%s",s[i]); for(i = 1; i <= n; i++) for(j = 0; j < m; j++) { if(s[i][j] == 'S') { map[i][j+1] = 0; x = i; y = j+1; } else if(s[i][j] == '.') map[i][j+1] = 0; else if(s[i][j] == '#') map[i][j+1] = 1; else if(s[i][j] == 'E') { map[i][j+1] = 0; a = i; b = j+1; } dp[i][j+1] = 0; } p = x; q = y; if(y == 1) f[x][y] = 2; else if(y == m) f[x][y] = 4; else if(x == 1) f[x][y] = 3; else if(x == n) f[x][y] = 1; while(1) { if(f[x][y] == 1) { if(map[x][y-1] == 0) { y -= 1; f[x][y] = 4; lcnt++; } else { if(map[x-1][y] == 0) { x -= 1; f[x][y] = 1; lcnt++; } else f[x][y] = 2;//向右转 } } else if(f[x][y] == 2) { if(map[x-1][y] == 0) { x -= 1; f[x][y] = 1; lcnt++; } else { if(map[x][y+1] == 0) { y += 1; f[x][y] = 2; lcnt ++; } else f[x][y] = 3; } } else if(f[x][y] == 3) { if(map[x][y+1] == 0) { y += 1; f[x][y] = 2; lcnt++; } else { if(map[x+1][y] == 0) { x += 1; f[x][y] = 3; lcnt++; } else f[x][y] = 4; } } else if(f[x][y] == 4) { if(map[x+1][y] == 0) { x += 1; f[x][y] = 3; lcnt++; } else { if(map[x][y-1] == 0) { y -= 1; f[x][y] = 4; lcnt++; } else f[x][y] = 1; } } if(a == x && b == y) break; } x = p, y = q; while(1) { if(f[x][y] == 1) { if(map[x][y+1] == 0) { y += 1; f[x][y] = 2; rcnt++; } else { if(map[x-1][y] == 0) { x -= 1; f[x][y] = 1; rcnt++; } else f[x][y] = 4;//向左转 } } else if(f[x][y] == 2) { if(map[x+1][y] == 0) { x += 1; f[x][y] = 3; rcnt++; } else { if(map[x][y+1] == 0) { y += 1; f[x][y] = 2; rcnt ++; } else f[x][y] = 1; } } else if(f[x][y] == 3) { if(map[x][y-1] == 0) { y -= 1; f[x][y] = 4; rcnt++; } else { if(map[x+1][y] == 0) { x += 1; f[x][y] = 3; rcnt++; } else f[x][y] = 2; } } else if(f[x][y] == 4) { if(map[x-1][y] == 0) { x -= 1; f[x][y] = 1; rcnt++; } else { if(map[x][y-1] == 0) { y -= 1; f[x][y] = 4; rcnt++; } else f[x][y] = 3; } } if(x == a && y == b) break; } t = 0 ; tt = 0; bfs[t].x1 = p; bfs[t].y1 = q; num[t] = 1; dp[bfs[t].x1][bfs[t].y1] = 1; while(1) { if(map[bfs[t].x1][bfs[t].y1-1] == 0 && dp[bfs[t].x1][bfs[t].y1-1] == 0)//左边 { tt++; bfs[tt].x1 = bfs[t].x1; bfs[tt].y1 = bfs[t].y1-1; num[tt] = num[t]+1; if(bfs[tt].x1 == a && bfs[tt].y1 == b) break; dp[bfs[tt].x1][bfs[tt].y1] = 1; } if(map[bfs[t].x1][bfs[t].y1+1] == 0 && dp[bfs[t].x1][bfs[t].y1+1] == 0)//右边 { tt++; bfs[tt].x1 = bfs[t].x1; bfs[tt].y1 = bfs[t].y1+1; num[tt] = num[t]+1; if(bfs[tt].x1 == a && bfs[tt].y1 == b) break; dp[bfs[tt].x1][bfs[tt].y1] = 1; } if(map[bfs[t].x1-1][bfs[t].y1] == 0 && dp[bfs[t].x1-1][bfs[t].y1] == 0)//上边 { tt++; bfs[tt].x1 = bfs[t].x1-1; bfs[tt].y1 = bfs[t].y1; num[tt] = num[t]+1; if(bfs[tt].x1 == a && bfs[tt].y1 == b) break; dp[bfs[tt].x1][bfs[tt].y1] = 1; } if(map[bfs[t].x1+1][bfs[t].y1] == 0 && dp[bfs[t].x1+1][bfs[t].y1] == 0)//下边 { tt++; bfs[tt].x1 = bfs[t].x1+1; bfs[tt].y1 = bfs[t].y1; num[tt] = num[t]+1; if(bfs[tt].x1 == a && bfs[tt].y1 == b) break; dp[bfs[tt].x1][bfs[tt].y1] = 1; } t++; } printf("%d %d %d\n",lcnt, rcnt, num[tt]); } return 0;}- poj 3083 Children of the Candy Corn
- poj 3083Children of the Candy Corn
- poj 3083 Children of the Candy Corn
- poj 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- poj 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- poj 3083 Children of the Candy Corn
- POJ 3083 - Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- poj 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- POJ 3083 Children of the Candy Corn
- poj 3083 Children of the Candy Corn
- hdu 4635 Strongly connected (强连通分量)
- 打印机外设通信 原理
- 前端经验
- HDU1114 Piggy-Bank
- hdu 1233 还是畅通工程
- POJ 3083 Children of the Candy Corn
- Linux下I2C驱动分析
- iOS给背景添加点击事件
- Android SDK更新以及ADT更新出现问题的解决办法
- C# Word 标题设置大纲级别和自动生成目录
- URAL1069
- 内存操作流
- 类 ArrayBlockingQueue<E>（一个由数组支持的有界阻塞队列。）
- jquery
