poj 3009 (深搜求最短路)

题目大意就是求在特定规则下的最短路，这个规则包含了消除障碍的操作。用BFS感觉选择消除障碍的时候不同路径会有影响，用DFS比较方便状态的还原（虽然效率比较低），因此这道题目采用DFS来写。

写的第一次提交代码是TLE，原因是忘记总步数>10就应该剪枝的条件。AC代码如下：

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 using namespace std;
 const int maxn = ;
 struct Point{
     int r, c;
     Point(int r=-, int c=-):r(r),c(c){}
 };
 Point s,t;
 int W,H;
 int G[maxn][maxn];
 const int dr[] = {,-,,};
 const int dc[] = {,,-,};
 int ans  = ;
 void dfs(int r,int c,int k){
     if(k >= ) return ;
     for(int i = ; i < ; i++){
         int nr = r;
         int nc = c;
         int is_walk = ;
         while(G[nr+dr[i]][nc+dc[i]]==){
             is_walk=;
             nr+=dr[i];
             nc+=dc[i];
             if(nr == t.r && nc == t.c){
                 ans = min(ans,k+);
                 return ;
             }
         }
         if(!is_walk)continue;
         if(G[nr+dr[i]][nc+dc[i]] == )continue ;
         if(G[nr+dr[i]][nc+dc[i]] == ){
             G[nr+dr[i]][nc+dc[i]] = ;
             dfs(nr,nc,k+);
             G[nr+dr[i]][nc+dc[i]] = ;
         }
     }
 }
 int main(){
     while(scanf("%d%d ", &W, &H) && (W || H)){
         ans  = ;
         for(int i = ; i <= H; i++){
             for(int j = ; j <= W; j++)
                 scanf("%d",&G[i][j]);
         }
         for(int i = ; i <= W+; i++)
             G[][i] = ,G[H+][i] = ;
         for(int i = ; i <= H+; i++)
             G[i][] = ,G[i][W+] = ;
         for(int i = ; i <= H; i++){
             for(int j = ; j <= W; j++){
                 if(G[i][j] == ){
                     s = Point(i,j);
                     G[i][j] = ;
                 }
                 else if (G[i][j] == ){
                     t = Point(i,j);
                     G[i][j] = ;
                 }
             }
         }
         dfs(s.r,s.c,);
         if(ans !=  && ans <= )printf("%d\n",ans);
         else printf("%d\n",-);
     }
     return ;
 }