hdu1010搜索+剪枝
来源:互联网 发布:dnf怎么总网络中断 编辑:程序博客网 时间:2024/05/17 23:48
Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 42078 Accepted Submission(s): 11382
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
DFS用回溯法框架来做
#include <iostream>
#include <cmath>
using namespace std;
int N,M,T;
int time_used;
char map[7][7];
int dirx[4]={-1,0,1,0};
int diry[4]={0,1,0,-1};
int sx,sy;
int ex,ey;
bool flag;
int Count;
void DFS(int x,int y)
{
int xx,yy;
if(flag)
return;
if((x == ex && y == ey)&& (time_used == T)) //达到条件退出
{
flag = true;
return;
}
if((T-time_used)%2 != (abs(x-ex)+abs(y-ey))%2) //奇偶减枝法
return;
if(abs(x-ex)+abs(y-ey) > T-time_used)
#include <cmath>
using namespace std;
int N,M,T;
int time_used;
char map[7][7];
int dirx[4]={-1,0,1,0};
int diry[4]={0,1,0,-1};
int sx,sy;
int ex,ey;
bool flag;
int Count;
void DFS(int x,int y)
{
int xx,yy;
if(flag)
return;
if((x == ex && y == ey)&& (time_used == T)) //达到条件退出
{
flag = true;
return;
}
if((T-time_used)%2 != (abs(x-ex)+abs(y-ey))%2) //奇偶减枝法
return;
if(abs(x-ex)+abs(y-ey) > T-time_used)
//当前点到终点的最短时间若比剩余的时间还长的话
return;
for(int i = 0; i < 4;i++) //向四个方向扩展
{
xx = x + dirx[i];
yy = y + diry[i];
if( (xx >= 0) && (xx < N) && (yy >= 0) && (yy <M) && map[xx][yy] != 'X')
{
map[x][y] = 'X';
time_used++;
DFS(xx,yy);
time_used--;
map[x][y] = '.';
}
}
}
int main()
{
while(cin>>N>>M>>T)
{
if(!N && !M && !T)
break;
Count = 0;
for(int i = 0;i < N;i++)
{
for(int j = 0;j < M;j++)
{
cin>>map[i][j];
if(map[i][j] == 'S')
{
sx = i;
sy = j;
}
if(map[i][j] == 'D')
{
ex = i;
ey = j;
}
else if(map[i][j] == '.')
{
Count++;
}
}
}
if(Count +1 < T) // 如果总共可以走的步数比给定时间少
{
cout<<"NO"<<endl;
continue;
}
time_used = 0;
flag = false;
DFS(sx,sy);
if(flag == true)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
return;
for(int i = 0; i < 4;i++) //向四个方向扩展
{
xx = x + dirx[i];
yy = y + diry[i];
if( (xx >= 0) && (xx < N) && (yy >= 0) && (yy <M) && map[xx][yy] != 'X')
{
map[x][y] = 'X';
time_used++;
DFS(xx,yy);
time_used--;
map[x][y] = '.';
}
}
}
int main()
{
while(cin>>N>>M>>T)
{
if(!N && !M && !T)
break;
Count = 0;
for(int i = 0;i < N;i++)
{
for(int j = 0;j < M;j++)
{
cin>>map[i][j];
if(map[i][j] == 'S')
{
sx = i;
sy = j;
}
if(map[i][j] == 'D')
{
ex = i;
ey = j;
}
else if(map[i][j] == '.')
{
Count++;
}
}
}
if(Count +1 < T) // 如果总共可以走的步数比给定时间少
{
cout<<"NO"<<endl;
continue;
}
time_used = 0;
flag = false;
DFS(sx,sy);
if(flag == true)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
扩展:
奇偶剪枝
- hdu1010搜索+剪枝
- HDU1010 奇偶剪枝 + 预先筛选 + DFS搜索
- HDU1010 - Tempter of the Bone (搜索+剪枝)
- hdu1010 Tempter of the Bone dfs搜索 剪枝
- hdu1010(深搜 +剪枝)
- hdu1010 DFS和剪枝
- hdu1010 深搜剪枝.
- HDU1010-奇偶剪枝(DFS)
- hdu1010 dfs+路径剪枝
- hdu1010(dfs+剪枝)
- HDU1010(回溯剪枝)
- hdu1010 dfs奇偶剪枝
- HDU1010奇偶剪枝DFS
- HDU1010 深搜+奇偶剪枝
- hdu1010 奇偶剪枝
- hdu1010 dfs+剪枝
- HDU1010 DFS+奇偶剪枝
- dfs+剪枝(hdu1010)
- Android异步下载网络图片(3)
- UBIFS设计简介
- 如何开启YII系统默认的DEBUG
- 对象的引用传递
- windows程序设计 例题解析 KEYVIEM1.C
- hdu1010搜索+剪枝
- OSI参考模型详解【原】
- 3DSMAX命令中英文对照表及快捷键大全
- SyncToy , 微软推出的一款免费的文件夹同步工具
- 移动支付三年内取代银行卡和 POS 机
- Win8 市场份额突破 1% XP 首次跌破 40%
- 给定数X,求X!的末尾连续零的个数
- 微软移动操作系统将迎来生死年
- 关于数组做为行参和实参