HDOJ1010.Tempter of the Bone

试题请参见: http://acm.hdu.edu.cn/showproblem.php?pid=1010

题目概述

The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

解题思路

迷宫问题, 深度优先搜索(回溯).
不过需要奇偶剪枝.

遇到的问题

一开始陷入了TLE, 之后又陷入了WA.

1. 需要引入奇偶剪枝, 也就是说, 当剩余步数( = 总步数 - 当前步数)为奇数时, 一定无法到达;
2. 一开始是纯暴力搜索, 并没有回溯. 引入了isVisited数组之后成功解决;
3. 一开始的思路是, 使用DFS计算出最小的步数, 然后再用奇偶剪枝, 但是事实证明, 这方法不可行(测试数据见下一节).

参考的测试数据

• http://acm.hdu.edu.cn/discuss/problem/post/reply.php?postid=7611&messageid=1&deep=0
• http://acm.hdu.edu.cn/discuss/problem/post/reply.php?postid=17995&messageid=1&deep=0

其中第2个数据就是为什么计算最小步数是不可行的原因.

源代码

#include <iostream>#include <cmath>const int MAX_SIZE = 10;bool isAccessible[MAX_SIZE][MAX_SIZE];bool isVisited[MAX_SIZE][MAX_SIZE];bool isExitReached = false;void traverseMap(int currentX, int currentY, int exitX, int exitY, int currentSteps, int n, int m, int t) {    if ( currentX < 0 || currentY < 0 || currentX >= n || currentY >= m ) {        return;    }    if ( isExitReached ) {        return;    }    if ( currentSteps == t && currentX == exitX && currentY == exitY ) {        isExitReached = true;        return;    }    // Branch Cutting    int restSteps = t - currentSteps - std::abs(currentX - exitX) - std::abs(currentY - exitY);    if ( restSteps < 0 || restSteps % 2 != 0 ) {        isExitReached = false;        return;    }    isVisited[currentX][currentY] = true;    if ( !isVisited[currentX][currentY - 1] && isAccessible[currentX][currentY - 1] ) {        traverseMap(currentX, currentY - 1, exitX, exitY, currentSteps + 1, n, m, t);    }    if ( !isVisited[currentX][currentY + 1] && isAccessible[currentX][currentY + 1] ) {        traverseMap(currentX, currentY + 1, exitX, exitY, currentSteps + 1, n, m, t);    }    if ( !isVisited[currentX - 1][currentY] && isAccessible[currentX - 1][currentY] ) {        traverseMap(currentX - 1, currentY, exitX, exitY, currentSteps + 1, n, m, t);    }    if ( !isVisited[currentX + 1][currentY] && isAccessible[currentX + 1][currentY] ) {        traverseMap(currentX + 1, currentY, exitX, exitY, currentSteps + 1, n, m, t);    }    isVisited[currentX][currentY] = false;}int main() {    int n = 0, m = 0, t = 0;    while ( std::cin >> n >> m >> t ) {        if ( n == 0 && m == 0 && t == 0 ) {            break;        }        isExitReached = false;        int startX = 0, startY = 0;        int exitX = 0, exitY = 0;        char mapPoint;        for ( int i = 0; i < n; ++ i ) {            for ( int j = 0; j < m; ++ j ) {                std::cin >> mapPoint;                if ( mapPoint == 'S' ) {                    isAccessible[i][j] = true;                    startX = i;                    startY = j;                } else if ( mapPoint == 'D' ) {                    isAccessible[i][j] = true;                    exitX = i;                    exitY = j;                } else if ( mapPoint == 'X' ) {                    isAccessible[i][j] = false;                } else if ( mapPoint == '.' ) {                    isAccessible[i][j] = true;                }            }        }        traverseMap(startX, startY, exitX, exitY, 0, n, m, t);        if ( isExitReached ) {            std::cout << "YES" << std::endl;        } else {            std::cout << "NO" << std::endl;        }    }    return 0;}