题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1010
Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 69549 Accepted Submission(s): 19126
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.
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.
'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
题目大意：S起点 D终点 X障碍物 .可走路 输入N,M,T 表示N行 M列 问你是否能在 T 时刻恰好到达终点D。（注意这里是刚刚好T步 多了少了都不行 而且不能重复走）
思路1：与上一题一样 hdu1242 Rescue 一样 记录所有能到终点的路径到时候找是否存在刚刚好一个长度正好为T的
注意：此思路的代码还能优化 读者有兴趣的可以试试
#include<iostream>
using namespace std;
#include<string.h>
#define max 205
char map[max][max];
long a[100000],step,sum,n,m,visited[max][max];
long directions[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
void DFS(int x,int y)
{
int i,mx,my;
if(map[x][y]=='D')
a[sum++]=step;
else if(map[x][y]!='X')
{
for(i=0;i<4;i++)
{
mx=x+directions[i][0];
my=y+directions[i][1];
if(map[mx][my]!='X'&&mx>=1&&mx<=n&&my>=1&&my<=m&&!visited[mx][my])//不是墙并且没走过
{
// if(map[x][y]=='x')
// step++;
step++;
visited[mx][my]=1;
DFS(mx,my); //所以关键要得到值的是递归的这一步 推导的时候让这个的下一个DFS就是到达朋友那点比较好理解为什么后面要还原
visited[mx][my]=0;//这一步的原因，上面DFS进行完后要将状态还原
step--; //下面这些都要还原
// if(map[x][y]=='x')
// step--;
}
}
}
}
int main()
{
long i,j,x,y,min;
int kk;
while(cin>>n>>m>>kk)
{
if(n==0&&m==0&&kk==0)break;
memset(visited,0,sizeof(visited));
sum=0;
step=0;
min=max;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
cin>>map[i][j];
if(map[i][j]=='S')//记录天使的地址
{
x=i;
y=j;
}
}
}
visited[x][y]=1;
DFS(x,y);
int flag=0;
for(i=0;i<sum;i++)
if(a[i]==kk){flag=1;break;}
if(flag==1)cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
思路2：
借用这里的，
奇偶剪枝：
是数据结构的搜索中剪枝的一种特殊小技巧。
现假设起点为(sx,sy)，终点为（ex,ey），给定t步恰好走到终点，
| s | ||||
| | | ||||
| | | ||||
| | | ||||
| + | — | — | — | e |
如图所示（“|”竖走，“—”横走，“+”转弯），易证abs(ex-sx)+abs(ey-sy)为此问题类中任意情况下，起点到终点的最短步数，记做step，此处step1=8；
| s | — | — | — | |
| — | — | + | ||
| | | + | |||
| | | ||||
| + | — | — | — | e |
如图，为一般情况下非最短路径的任意走法举例，step2=14；
step2-step1=6，偏移路径为6，偶数（易证）；
故，若t-[abs(ex-sx)+abs(ey-sy)]结果为非偶数（奇数），则无法在t步恰好到达；
返回，false；
反之亦反。
280kb,156ms;
#include<iostream>
#include<cstring>
#define N 10
using namespace std;
int n,m,t,end_i,end_j;
bool visited[N][N],flag,ans;
char map[N][N];
int a[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int abs(int a,int b)//用cmath也可以
{//
if(a<b) return b-a;//
else return a-b;//
}//
void DFS(int i,int j,int c)
{
if(flag) return ;
if(c>t) return ;
if(i<0||i>=n||j<0||j>=m) {return ;}
if(map[i][j]=='D'&&c==t) {flag=ans=true; return ;}
int temp=abs(i-end_i)+abs(j-end_j);
temp=t-temp-c; //t扣掉还要走的最短步temp 和 已经走过的 c 如果这些步还是奇数直接不满足
if(temp&1) return ;//奇偶剪枝 奇数return
for(int k=0;k<4;k++)
if(!visited[i+a[k][0]][j+a[k][1]]&&map[i+a[k][0]][j+a[k][1]]!='X') //开始进行各个方向的探索 记得回溯，取消之前走的状态
{
visited[i+a[k][0]][j+a[k][1]]=true;
DFS(i+a[k][0],j+a[k][1],c+1);
visited[i+a[k][0]][j+a[k][1]]=false;
}
}
int main()
{
int i,j,x,y,k;
while(cin>>m>>n>>t&&(m||n||t))
{
memset(visited,false,sizeof(visited));
k=0;//记录障碍物的数量
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
cin>>map[i][j];
if(map[i][j]=='S')
{
x=i;y=j;
visited[i][j]=true;
}
if(map[i][j]=='D')
{
end_i=i;end_j=j;
}
if(map[i][j]=='X')k++;
}
}
ans=flag=false;
if(n*m-k-1>=t) DFS(x,y,0);
if(ans) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
与上面的思路一样，代码写长了 不过时间变短了 也就是那个for(0->3)这里不一样
280 KB 78 ms
#include<iostream>
#include<cstring>
#define N 10
using namespace std;
int n,m,t,end_i,end_j;
bool visited[N][N],flag,ans;
char map[N][N];
int abs(int a,int b)
{
if(a<b) return b-a;
else return a-b;
}
void DFS(int i,int j,int c)
{
if(flag) return ;
if(c>t) return ;
if(i<0||i>=n||j<0||j>=m) {return ;}
if(map[i][j]=='D'&&c==t) {flag=ans=true; return ;}
int temp=abs(i-end_i)+abs(j-end_j);
temp=t-temp-c;
if(temp&1) return ;//奇偶剪枝
if(!visited[i-1][j]&&map[i-1][j]!='X')
{
visited[i-1][j]=true;
DFS(i-1,j,c+1);
visited[i-1][j]=false;
}
if(!visited[i+1][j]&&map[i+1][j]!='X')
{
visited[i+1][j]=true;
DFS(i+1,j,c+1);
visited[i+1][j]=false;
}
if(!visited[i][j-1]&&map[i][j-1]!='X')
{
visited[i][j-1]=true;
DFS(i,j-1,c+1);
visited[i][j-1]=false;
}
if(!visited[i][j+1]&&map[i][j+1]!='X')
{
visited[i][j+1]=true;
DFS(i,j+1,c+1);
visited[i][j+1]=false;
}
}
int main()
{
int i,j,x,y,k;
while(cin>>m>>n>>t&&(m||n||t))
{
memset(visited,false,sizeof(visited));
k=0;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
{
cin>>map[i][j];
if(map[i][j]=='S')
{
x=i;y=j;
visited[i][j]=true;
}
if(map[i][j]=='D')
{
end_i=i;end_j=j;
}
if(map[i][j]=='X')k++;
}
}
ans=flag=false;
if(n*m-k-1>=t) DFS(x,y,0);
if(ans) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。