POJ 2935 BFS

Give the matrix of 6*6, the starting point, the end point, altogether three walls, the wall will not intersect.

Minimum step to the end of the starting point to ensure a solution

Whether there is a wall that corresponds to each move judgment

#include "stdio.h" #include "string.h" #include "queue" using namespace Std;const int dir[4][2]={{1,0},{-1,0},{0,1},{0,- 1}};int wall[10][10][10][10];int s_x,s_y,e_x,e_y;int used[10][10];struct node{int x, y;};    void pri (int x,int y) {if (used[x][y]==0) return;        if (used[x][y]==1) {pri (x-1,y);    printf ("S");        } if (used[x][y]==2) {pri (x+1,y);    printf ("N");        } if (used[x][y]==3) {pri (x,y-1);    printf ("E");        } if (used[x][y]==4) {pri (x,y+1);    printf ("W");    }}void BFs () {queue<node>q;    Node Cur,next;    int i;    cur.x=s_x;    cur.y=s_y;    Q.push (cur);    memset (used,-1,sizeof (used));    used[cur.x][cur.y]=0;        while (!q.empty ()) {Cur=q.front ();        Q.pop ();            for (i=0;i<4;i++) {next.x=cur.x+dir[i][0];            NEXT.Y=CUR.Y+DIR[I][1];            if (next.x<=0 | | next.x>6 | | next.y<=0 | | next.y>6) continue; if (i==0 &&            Wall[cur.x][cur.y-1][cur.x][cur.y]==1) continue;            if (i==1 && wall[next.x][next.y-1][next.x][next.y]==1) continue;            if (i==2 && wall[cur.x-1][cur.y][cur.x][cur.y]==1) continue;            if (i==3 && wall[next.x-1][next.y][next.x][next.y]==1) continue;            if (used[next.x][next.y]!=-1) continue;            used[next.x][next.y]=i+1;            Q.push (next);                if (next.x==e_x && next.y==e_y) {pri (NEXT.X,NEXT.Y);                printf ("\ n");            return;    }}}}int Main () {int i,j,x1,x2,y1,y2,temp;        while (scanf ("%d%d", &s_y,&s_x)!=eof) {if (s_x+s_y==0) break;        scanf ("%d%d", &e_y,&e_x);        memset (wall,0,sizeof (wall));            for (i=1;i<=3;i++) {scanf ("%d%d%d%d", &y1,&x1,&y2,&x2);                if (x1==x2) {if (y1>y2) {temp=y1; y1=y2; y2=temp;} FoR (j=y1;j<y2;j++) wall[x1][j][x1][j+1]=1;                } else {if (x1>x2) {temp=x1; x1=x2; x2=temp;}            for (j=x1;j<x2;j++) wall[j][y1][j+1][y1]=1;    }} BFS (); } return 0;}

POJ 2935 BFS