hdu 1010 Tempter of the Bone(dfs剪枝+回溯)

最新推荐文章于 2021-10-21 18:26:48 发布

jxust_xiaoxiong

最新推荐文章于 2021-10-21 18:26:48 发布

阅读量479

点赞数

CC 4.0 BY-SA版权

分类专栏：搜索

本文链接：https://blog.youkuaiyun.com/Love_Lydia/article/details/38413257

搜索专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一个迷宫逃脱问题，通过深度优先搜索（DFS）解决小狗如何在限定时间内找到出口的问题。考虑了迷宫布局、时间限制及地面消失等条件，采用剪枝策略优化搜索过程。

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 70265 Accepted Submission(s): 19362

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

题意：一只小狗在迷宫里闻到了一块骨头的香味，要刚好T时刻到达才能得到，走过的格子会消失。

dfs剪枝 1.从a出发到b点的方格问题中a到b的各条路径与a到b的最短路径同奇偶
        2.格子数小于时间T就不可能到达
dfs回溯
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
using namespace std;
int n,m,t,ans,q,p;
int dir[][2]={1,0,-1,0,0,1,0,-1};//方向
char map[10][10];
void dfs(int x ,int y,int T)
{
    if(x==q&&y==p&&T==t)
    {
        ans=1;
        return ;
    }
    int s=t-fabs(x-q)-fabs(y-p)-T;//剪枝1
    if(s<0||s%2)
        return ;
    for(int i=0;i<4;i++)
    {
        int a=x+dir[i][0];
        int b=y+dir[i][1];
        if(a<1||b<1||a>n||b>m||map[a][b]=='X')
            continue;
        map[a][b]='X';
        dfs(a,b,T+1);
        if(ans)
            return ;
        map[a][b]='.';//回溯
    }
    return ;
}
int main()
{
    while(cin>>n>>m>>t,n+m+t)
    {
        int x,y,count=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='S')
                {
                    x=i;
                    y=j;
                }
                if(map[i][j]=='D')
                {
                    q=i;
                    p=j;
                }
                if(map[i][j]=='X')
                    count++;
            }
        }
        if(n*m-count<=t)//剪枝2
        {
            cout<<"NO"<<endl;
            continue;
        }
        ans=0;
        map[x][y]='X';
        dfs(x,y,0);
        if(ans==1)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}