HDU 1254 push box

An interesting BFS + DFS. Attached data.

Today's task was to carry multiple backpacks. In order to help others find wa points, I am stuck in this question.

Finally, I thought about a group of data that I did not have... No idea... I changed my mind.

The correct idea is to find the minimum BFS starting from the box. When moving each time, DFS determines whether a person can move to the back of the box.

At the beginning, I wrote a BFs, and all the data had passed. This group cannot pass

17 40 0 0 00 0 1 00 2 0 31 4 1 01 0 1 01 0 1 01 0 0 0

Actually, the answer is 2.

What I wrote is the shortest step of the box and the shortest step of the box. To wa.

Then I changed my mind. Note that the status should be four-dimensional, because the box may need to be pushed first and then pushed back.

3 6
1 1 1 1 1
0 0 0 1 0 0
0 0 2 4 0 3

For example.

Actually 5.

#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<queue>#include<map>#include<stack>#include<iostream>#include<list>#include<set>#include<cmath>#define INF 0x7fffffff#define eps 1e-6using namespace std;int g[8][8];int xx[]={1,-1,0,0};int yy[]={0,0,-1,1};int n,m;struct node{    int x,y,lv;};node box,man,over;bool vis[8][8];bool ok;void dfs(int x1,int y1,int x2,int y2){    if(ok)return;    if(x1==x2&&y1==y2)    {        ok=1;        return;    }    int x,y;    for(int k=0;k<4;k++)    {        x=x1+xx[k];        y=y1+yy[k];        if(x>=n||x<0||y>=m||y<0||g[x][y]==1||vis[x][y])            continue;        vis[x][y]=1;        dfs(x,y,x2,y2);    }}int bfs(){    bool way[8][8][8][8];    queue<node>q;    queue<node>human;    memset(way,0,sizeof(way));    q.push(box);    human.push(man);    way[box.x][box.y][man.x][man.y]=1;    node tmp,now,ma;    while(!q.empty())    {        tmp=q.front();q.pop();        ma=human.front();human.pop();        if(tmp.x==over.x&&tmp.y==over.y)return tmp.lv;        for(int k=0;k<4;k++)        {            int x=tmp.x+xx[k];            int y=tmp.y+yy[k];            if(x>=n||x<0||y>=m||y<0||g[x][y]==1||way[x][y][tmp.x][tmp.y])            continue;            int rex=tmp.x-xx[k];            int rey=tmp.y-yy[k];            if(rex>=n||rex<0||rey>=m||rey<0||g[rex][rey]==1)            continue;            memset(vis,0,sizeof(vis));            ok=0;            vis[tmp.x][tmp.y]=1;            //printf("box %d %d -> %d %d\n",tmp.x,tmp.y,x,y);            //printf("human %d %d -> %d %d\n",ma.x,ma.y,rex,rey);            //system("pause");            dfs(ma.x,ma.y,rex,rey);            if(!ok)            continue;            now.x=x,now.y=y,now.lv=tmp.lv+1;            way[x][y][tmp.x][tmp.y]=1;            q.push(now);            human.push(tmp);        }    }    return -1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=0;i<n;i++)        for(int j=0;j<m;j++)        {            scanf("%d",&g[i][j]);            if(g[i][j]==2)                box.x=i,box.y=j,box.lv=0;            else if(g[i][j]==3)                over.x=i,over.y=j;            else if(g[i][j]==4)                man.x=i,man.y=j;        }        int tmp=bfs();        printf("%d\n",tmp);    }}