关于 BFS 中 打印路径的问题

打印路径可以用栈的方式， 也可以用递归。下面看看用递归做的吧还是NYOJ 最少步数那道题

#include <iostream>
#include<queue>
#include<cstring>
#include<cstdio>
const int dx[4] = {0, 0,-1, 1};
const int dy[4] = {-1,1, 0, 0};
struct node
{
int x, y;
int step;
};
int x1, y1, x2, y2;
node now, next;
node S,E;

using namespace std;
bool vis[10][10];
node qq[10][10];
int map[9][9] = {1,1,1,1,1,1,1,1,1
, 1,0,0,1,0,0,1,0,1
,1,0,0,1,1,0,0,0,1
,1,0,1,0,1,1,0,1,1
,1,0,0,0,0,1,0,0,1
,1,1,0,1,0,1,0,0,1
,1,1,0,1,0,1,0,0,1
,1,1,0,1,0,0,0,0,1
,1,1,1,1,1,1,1,1,1};
int bfs()
{
if(S.x == E.x && S.y == E.y)
return 0;
memset(vis, false,sizeof(vis));
queue<node>q;
q.push(S);
vis[S.x][S.y] = true;
while(!q.empty())
{
now = q.front();
q.pop();
for(int i=0; i<4; i++)
{
next.x = now.x + dx[i];
next.y = now.y + dy[i];
next.step = now.step + 1;
if(map[next.x][next.y] == 1||next.x > 9 || next.x<0 || next.y>9 || next.y<0 || vis[next.x][next.y])
continue;
vis[next.x][next.y] = true;
qq[next.x][next.y].x = now.x;
qq[next.x][next.y].y = now.y;
q.push(next);
if(next.x == E.x && next.y == E.y)
return next.step;
}
}
return -1;

}

void pri(int sx, int sy) {
if( sx==x1 && sy==y1 ) {
printf("(%d, %d)\n", sx, sy);
return ;
}
pri(qq[sx][sy].x, qq[sx][sy].y);
printf("(%d, %d)\n", sx, sy);
}
int main()
{
int T;

cin >> T;
while(T--)
{
cin >> x1 >> y1;
cin >> x2 >> y2;
S.x = x1;
S.y = y1;
S.step = 0;
E.x = x2;
E.y = y2;

int ans = bfs();
if(ans >= 0)
{
cout << ans << endl;
pri(x2, y2);
}

}
return 0;
}

下面是用栈实现的：

#include <iostream>#include<queue>#include<cstring>#include<stack>const int dx[4] = {0, 0,-1, 1};const int dy[4] = {-1,1, 0, 0};struct node{int x, y;int step;};int x1, y1, x2, y2;node now, next;node S,E;node Pre[10][10];using namespace std;bool vis[10][10];int map[9][9] = {1,1,1,1,1,1,1,1,1, 1,0,0,1,0,0,1,0,1,1,0,0,1,1,0,0,0,1,1,0,1,0,1,1,0,1,1,1,0,0,0,0,1,0,0,1,1,1,0,1,0,1,0,0,1,1,1,0,1,0,1,0,0,1,1,1,0,1,0,0,0,0,1,1,1,1,1,1,1,1,1,1};int bfs(){if(S.x == E.x && S.y == E.y)return 0;memset(vis, false,sizeof(vis));queue<node>q;q.push(S);vis[S.x][S.y] = true;while(!q.empty()){now = q.front();q.pop();for(int i=0; i<4; i++){next.x = now.x + dx[i];next.y = now.y + dy[i];next.step = now.step + 1;if(map[next.x][next.y] == 1||next.x > 9 || next.x<0 || next.y>9 || next.y<0 || vis[next.x][next.y])continue;vis[next.x][next.y] = true;Pre[next.x][next.y] = now;q.push(next);if(next.x == E.x && next.y == E.y)return next.step;}}return -1;}void PrintPath(int row,int column){//输出最短路径node temp;         //保存位置stack<node> s;     //保存路径序列temp.x = row;temp.y = column;while(temp.x != x1 || temp.y != y1){s.push(temp);temp = Pre[temp.x][temp.y];}cout<<"(" << x1 << "," << y1 << ")" << endl;while(!s.empty()){temp = s.top();cout<<"("<<temp.x<<","<<temp.y<<")" << endl;s.pop();}cout<<endl;}int main(){int T;cin >> T;while(T--){cin >> x1 >> y1;cin >> x2 >> y2;S.x = x1;S.y = y1;S.step = 0;E.x = x2;E.y = y2;int ans = bfs();if(ans >= 0){cout << ans << endl;PrintPath(x2, y2);}}return 0;}/*23 1  5 712(3,1)(4,1)(4,2)(4,3)(4,4)(5,4)(6,4)(7,4)(7,5)(7,6)(7,7)(6,7)(5,7)3 1 6 711(3,1)(4,1)(4,2)(4,3)(4,4)(5,4)(6,4)(7,4)(7,5)(7,6)(7,7)(6,7)Process returned 0 (0x0)   execution time : 24.583 sPress any key to continue.*/

再做做下面这道题， 越来越发现应该学会融会贯通啊！题目描述：一个网格迷宫由n行m列的单元格组成，每个单元格要么是空地（用1表示），要么是障碍物（用0来表示）。你的任务是找一条从起点到终点的最短步数和移动序列，其中UDLR表示上下左右操作。任何时候都不能在障碍物格子中，也不能走到迷宫之外。起点和终点保证都是空地。n，m<100。样例输入：6 5110111011110100101111110111111样例输出：9DDDDDRRRR解题思路：较上一个文章来说，只是多了输出路径，这里既然题目没规定入口出口，为了方便起见，我人为的把入口和出口规定为左上角和右下角。记录路径即需要知道终点是由那个点走到的，而这个点的上一个点也需要知道……所以就记录下每个点的父节点和相应的操作，然而准备输出路径的时候又遇到了一个问题：从出口一步步向上寻找，是一个倒着的，怎么才能正着输出呢？先进后出这个特点是栈对应的特点，而递归类似栈，所以利用递归的方式，达到按顺序输出的目的。代码：[cpp] viewplain copy#include<cstdio>#include<queue>#include<iostream>using namespace std;char mmap[10][10]={};int vis[10][10]={};int n,m;int xx[4]={1,0,0,-1};int yy[4]={0,1,-1,0};struct node{int x,y;int t;};struct father{int x,y;//当前格子的父节点坐标char cz;//由什么操作到达的这个格子};queue<node> q;node s,f;father lj[10][10];//记录路径int bfs(){int i,j;s.x=0;s.y=0;s.t=0;f.x=n-1;f.y=m-1;q.push(s);lj[s.x][s.y].x=1000;//因为m，n<=100lj[s.x][s.y].y=1000;lj[s.x][s.y].cz=0;vis[s.x][s.y]=1;//标为已经访问过while(!q.empty()){node now=q.front();q.pop();for(i=0;i<4;++i){node New;New.x=now.x+xx[i];New.y=now.y+yy[i];New.t=now.t+1;if(New.x<0||New.y<0||New.x>=n||New.y>=m||vis[New.x][New.y]||mmap[New.x][New.y]=='0')//下标越界或者访问过或者是障碍物continue;q.push(New);lj[New.x][New.y].x=now.x;lj[New.x][New.y].y=now.y;if(i==0) lj[New.x][New.y].cz='D';else if(i==1) lj[New.x][New.y].cz='R';else if(i==2) lj[New.x][New.y].cz='L';else if(i==3) lj[New.x][New.y].cz='U';vis[New.x][New.y]=1;if(New.x==f.x&&New.y==f.y) return New.t;//到达终点}}return -1;}void dfs(int x, int y){if(x==0&&y==0) return;//找到父节点是起点的格子了elsedfs(lj[x][y].x,lj[x][y].y);printf("%c",lj[x][y].cz);}int main(){int i,j,ans;scanf("%d%d",&n,&m);for(i=0;i<n;++i){scanf("%s",mmap[i]);}ans=bfs();if(ans<0) printf("error");else{printf("%d\n",ans);dfs(n-1,m-1);//从终点开始找他的父节点}return 0;}