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 = 22;
struct Point{
    int r, c;
    Point(int r=-1, int c=-1):r(r),c(c){}
};
Point s,t;
int W,H;
int G[maxn][maxn];
const int dr[] = {1,-1,0,0};
const int dc[] = {0,0,-1,1};
int ans  = 100000;
void dfs(int r,int c,int k){
    if(k >= 10) return ;
    for(int i = 0; i < 4; i++){
        int nr = r;
        int nc = c;
        int is_walk = 0;
        while(G[nr+dr[i]][nc+dc[i]]==0){
            is_walk=1;
            nr+=dr[i];
            nc+=dc[i];
            if(nr == t.r && nc == t.c){
                ans = min(ans,k+1);
                return ;
            }
        }
        if(!is_walk)continue;
        if(G[nr+dr[i]][nc+dc[i]] == 4)continue ;
        if(G[nr+dr[i]][nc+dc[i]] == 1){
            G[nr+dr[i]][nc+dc[i]] = 0;
            dfs(nr,nc,k+1);
            G[nr+dr[i]][nc+dc[i]] = 1;
        }
    }
}
int main(){
    while(scanf("%d%d ", &W, &H) && (W || H)){
        ans  = 100000;
        for(int i = 1; i <= H; i++){
            for(int j = 1; j <= W; j++)
                scanf("%d",&G[i][j]);
        }
        for(int i = 0; i <= W+1; i++)
            G[0][i] = 4,G[H+1][i] = 4;
        for(int i = 0; i <= H+1; i++)
            G[i][0] = 4,G[i][W+1] = 4;
        for(int i = 1; i <= H; i++){
            for(int j = 1; j <= W; j++){
                if(G[i][j] == 2){
                    s = Point(i,j);
                    G[i][j] = 0;
                }
                else if (G[i][j] == 3){
                    t = Point(i,j);
                    G[i][j] = 0;
                }
            }
        }
        dfs(s.r,s.c,0);
        if(ans != 100000 && ans <= 10)printf("%d
",ans);
        else printf("%d
",-1);
    }
    return 0;
}