hdu

奇偶剪枝

题目链接
Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 104098 Accepted Submission(s): 28268

Problem Description
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.

Input
The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

‘X’: a block of wall, which the doggie cannot enter;
‘S’: the start point of the doggie;
‘D’: the Door; or
‘.’: an empty block.

The input is terminated with three 0’s. This test case is not to be processed.

Output
For each test case, print in one line “YES” if the doggie can survive, or “NO” otherwise.

Sample Input
4 4 5
S.X.
..X.
..XD
….
3 4 5
S.X.
..X.
…D
0 0 0

Sample Output
NO
YES

一个N*M迷宫,从’S’开始经过’.’使用T步到达’D’,每走过一点’.’就将其设成’X’表示不能通过.判断给定条件下能否到达.

可以利用DFS,不过会超时.引入奇偶剪枝概念.
从方格中行走,如A点到B点,则A到B的任一路径的步数要么全是奇数,要么全是偶数.这就是奇偶剪枝的原理.
除此之外还有显而易见的剪枝技巧:

  1. T不能小于A到B的最短路径步数.
  2. ‘.’的个数不能小于T-1.
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<cstdlib>
#define MAX 10

using namespace std;

//上下左右4个方向
int dx[4] = {-1, 0, 1, 0}, dy[4]= {0,1,0,-1};

char maze[MAX][MAX];
int N, M, sx,sy,ex,ey,step;
bool escape;

/**
* @name fun DFS,奇偶剪枝
* 到达m[i][j]时共走了t步
*/
void fun(char (*m)[MAX], int i, int j, int t)
{
    if(escape)return;
    int deta = step - t - abs(i+j-ex-ey); //奇偶剪枝
    if(deta<0 || deta & 1)return;
    if(i==ex && j==ey)
    {
        if(step==t)
        {
            escape = true;
        }
        return;
    }
    int nx, ny;
    for(int x = 0; x<4; x++)
    {
        nx = i+dx[x], ny = j+dy[x];
        if(0<=nx&&nx<N && 0<=ny&&ny<M && m[nx][ny]!='X')
        {
            m[nx][ny]='X';
            fun(m, nx, ny, t+1);
            m[nx][ny]='.';
        }
    }
}

int main()
{
    int dots;

    while(scanf("%d %d %d", &N, &M, &step))
    {
        if(N==0 && M==0 && step==0)break;
        escape = false;
        dots = 0;
        for(int i = 0; i < N; i++)
        {
            scanf("%s", maze[i]);
            for(int j=0; j<M; j++)
            {
                if(maze[i][j]=='S')
                {
                    sx = i;
                    sy = j;
                    maze[i][j]='X';
                }
                else if(maze[i][j]=='D')
                {
                    ex = i;
                    ey = j;
                }
                else if(maze[i][j]=='.')
                {
                    dots++;
                }
            }
        }
        if(abs(ex+ey-sx-sy) > step || dots < step-1) //step小于最短路径或'.'的个数小于step-1个, 特别注意是step-1
        {
            printf("NO\n");
            continue;
        }
        fun(maze, sx, sy, 0);
        if(escape)
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
    return 0;
}