首页 > 代码库 > hdu 1010 Tempter of the Bone （DFS+剪枝）

hdu 1010 Tempter of the Bone （DFS+剪枝）

2024-07-12 16:51:51   221人阅读

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 68206    Accepted Submission(s): 18719


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 5S.X...X...XD....3 4 5S.X...X....D0 0 0
 

 

Sample Output
NOYES
 

 

 

           题意：就是狗要找到一条路在规定的步数正好走到门口，每走一步是1s。

           这道题写了几次WA了，后面发现自己还是不太会剪枝，学了网上的代码懂了很多，但还是WA，代码甚至完全一样还是WA，查了半天没有找到不一样的，后面发现是输入问题，我用的是scanf("%c", &map[i][j]);而且我也处理了缓冲区问题，后面把输入该了就过了，下次遇到这个字符输入还是首选cin吧。

 

贴代码：

 

#include <stdio.h>#include <stdlib.h>#include <iostream>using namespace std;char map[10][10];int n, m, mark, step, finishx, finishy;int dir[4][2]={0, -1, 0, 1, 1, 0, -1, 0};void DFS(int startx, int starty, int counting){    int temp, x, y;    if(startx<1 || startx>n || starty<1 || starty>m)        return ;    if(startx == finishx && starty == finishy && counting == step)    {        mark = 1;        return ;    }    temp = (step-counting)-abs(startx-finishx)-abs(starty-finishy);    //判断剩余的步数是否为偶数或大于0    if(temp<0 || temp&1)        return ;    for(int i = 0; i<4; i++)    {         x = startx+dir[i][0];   y = starty+dir[i][1];        if(map[x][y] != ‘X‘)        {            map[x][y] = ‘X‘;            DFS(x, y, counting+1);            if(mark)                return;            map[x][y] = ‘.‘;        }    }    return;}int main(){    int startx, starty, wall;    while(scanf("%d%d%d", &n, &m, &step)!=EOF)    {        if(n==0 && m==0 && step==0)            break;        mark = 0;         wall = 0;        for(int i = 1; i<=n; i++)        {            //getchar();            for(int j = 1; j<=m; j++)            {                //scanf("%c", &map[i][j]);                cin>>map[i][j];           //注意这里一个坑，用cin吧                if(map[i][j] == ‘S‘)                {                    startx = i;        starty = j;                }                if(map[i][j] == ‘D‘)                {                    finishx = i;      finishy = j;                }                if(map[i][j] == ‘X‘)                    wall++;            }        }        if(m*n-wall <= step)        {            printf("NO\n");            continue;        }        map[startx][starty] = ‘X‘;        DFS(startx, starty, 0);        if(mark)            printf("YES\n");        else            printf("NO\n");    }    return 0;}


联系
我们