标签：深搜的方式
Description
On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves.
Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again.
Fig. 1: Example of board (S: start, G: goal)
The movement of the stone obeys the following rules:
Fig. 2: Stone movements
Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.
With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).
Fig. 3: The solution for Fig. D-1 and the final board configuration
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.
Each dataset is formatted as follows.
the width(=w) and the height(=h) of the board
First row of the board
...
h-th row of the board
The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.
Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.
0 vacant square 1 block 2 start position 3 goal position
The dataset for Fig. D-1 is as follows:
6 6
1 0 0 2 1 0
1 1 0 0 0 0
0 0 0 0 0 3
0 0 0 0 0 0
1 0 0 0 0 1
0 1 1 1 1 1
Output
For each dataset, print a line having a decimal integer indicating the minimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.
Sample Input
2 1 3 2 6 6 1 0 0 2 1 0 1 1 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 6 1 1 1 2 1 1 3 6 1 1 0 2 1 1 3 12 1 2 0 1 1 1 1 1 1 1 1 1 3 13 1 2 0 1 1 1 1 1 1 1 1 1 1 3 0 0
Sample Output
1 4 -1 4 10 -1
题意：给定一个由0,1,2，3 组成的矩阵，0 表示p平地，可以走通，1表示障碍物，不可走，2 表示移动物，可以击碎障碍物并停留在障碍物之前，3是出口，题意要求让你判断找到在有限的步骤（10步）内到达出口；
思路：主要分为两步1.移动到障碍物之前，超出界就说明这个方向不能走2.如何进行下个点的搜索
AC代码：
#include<iostream>
#include<cstdio>
using namespace std;
int w,h;
int sx,sy;
int ex,ey;
int sum;
int gra[21][21];
int xx[]={0,0,1,-1};
int yy[]={1,-1,0,0};
int isok(int x,int y){
if(x>=0&&x<h&&y>=0&&y<w){
return 1;
}
return 0;
}//越界函数的判断
void dfs(int tol,int x,int y){
int i;
if(tol>=10){
return;
}//走的步数大于10 直接返回
for(i=0;i<4;i++){//四个方向
int a=x+xx[i],b=y+yy[i];
if((gra[a][b]==0)||(gra[a][b]==3)){
while(gra[a][b]==0){//移动到障碍物之前的方法
a+=xx[i];
b+=yy[i];
}
if(isok(a,b)){//没有越界
if(gra[a][b]==3){
if(tol+1<sum){
sum=tol+1;
}
}//正好到达目标处
if(gra[a][b]==1){
gra[a][b]=0;
dfs(tol+1,a-xx[i],b-yy[i]);
gra[a][b]=1;
}
}
}
}
}
int main(){
while(~scanf("%d%d",&w,&h)&&w&&h){
sum=11;
for(int i=0;i<h;i++){
for(int j=0;j<w;j++){
scanf("%d",&gra[i][j]);
if(gra[i][j]==2){
gra[i][j]=0;
sx=i;
sy=j;
}
if(gra[i][j]==3){
ex=i;
ex=j;
}
}
}
dfs(0,sx,sy);
if(sum==11){
printf("-1\n");
}
else{
printf("%d\n",sum);
}
}
return 0;
}
注意：之前想到用if来移动，相当麻烦，还是要多想想，不能急的敲代码，多做做题。。。。。。
版权声明：本文为博主原创文章，未经博主允许不得转载。
标签：深搜的方式
原文地址：http://blog.csdn.net/qiuxueming_csdn/article/details/48102867
