Bloxorz I (poj3322) (BFS)

【题目描述】

	 It's a game about rolling a box to a specific position on a special plane. Precisely, the plane, which is composed of several unit cells, is a rectangle shaped area. And the box, consisting of two perfectly aligned unit cube, may either lies down and occupies two neighbouring cells or stands up and occupies one single cell. One may move the box by picking one of the four edges of the box on the ground and rolling the box 90 degrees around that edge, which is counted as one move. There are three kinds of cells, rigid cells, easily broken cells and empty cells. A rigid cell can support full weight of the box, so it can be either one of the two cells that the box lies on or the cell that the box fully stands on. A easily broken cells can only support half the weight of the box, so it cannot be the only cell that the box stands on. An empty cell cannot support anything, so there cannot be any part of the box on that cell. The target of the game is to roll the box standing onto the only target cell on the plane with minimum moves.

【算法】

	貌似没啥算法，就是BFS，但是写起来很烦，%lyd，大佬的代码为什么就这么清晰明了。。。。

【题目链接】

Bloxorz I

【代码】

#include <stdio.h>

#include <queue>

using namespace std;

struct rec{ int x,y,state; }st,ed;

char s[510][510];

int m,n,d[510][510][3];

queue<rec> q;

const int dx[]={0,0,-1,1},dy[]={-1,1,0,0};

bool valid(int x,int y) {

    return x>=1&&x<=m&&y>=1&&y<=n;

}

void parse_st_ed() {

    for(int i=1;i<=m;i++) {

        for(int j=1;j<=n;j++) {

            if(s[i][j]=='O') {

                ed.x=i,ed.y=j,s[i][j]='.';

            }else if(s[i][j]=='X') {

                for(int k=0;k<4;k++) {

                    int x=i+dx[k],y=j+dy[k];

                    if(valid(x,y)&&s[x][y]=='X') {

                        st.x=min(x,i),st.y=min(y,j);

                        st.state=x==i?1:2;

                        s[i][j]=s[x][y]='.';

                        break;

                    }

                }

                if(s[i][j]=='X') st.x=i,st.y=j,st.state=0;

            }

        }

    }

}

const int next_x[3][4]={ {0,0,-2,1},{0,0,-1,1},{0,0,-1,2} };

const int next_y[3][4]={ {-2,1,0,0},{-1,2,0,0},{-1,1,0,0} };

const int next_state[3][4]={ {1,1,2,2},{0,0,1,1},{2,2,0,0} };

bool valid(rec k) {

    if(!valid(k.x,k.y)) return 0;

    if(s[k.x][k.y]=='#') return 0;

    if(k.state==0&&s[k.x][k.y]=='E') return 0;

    if(k.state==1&&s[k.x][k.y+1]=='#') return 0;

    if(k.state==2&&s[k.x+1][k.y]=='#') return 0;

    return 1;

}

int bfs() {

    while(q.size()) q.pop();

    for(int i=1;i<=m;i++)

        for(int j=1;j<=n;j++)

            d[i][j][0]=d[i][j][1]=d[i][j][2]=-1;

    d[st.x][st.y][st.state]=0;

    q.push(st);

    while(q.size()) {

        rec now=q.front(); q.pop();

        for(int i=0;i<4;i++) {

            rec next;

            next.x=now.x+next_x[now.state][i];

            next.y=now.y+next_y[now.state][i];

            next.state=next_state[now.state][i];

            if(!valid(next)) continue;

            if(d[next.x][next.y][next.state]==-1) {

                d[next.x][next.y][next.state]

                    =d[now.x][now.y][now.state]+1;

                if(next.x==ed.x&&next.y==ed.y&&!next.state) return d[next.x][next.y][0];

                q.push(next);

            }

        }

    }

    return -1;

}

int main() {

    while(~scanf("%d%d",&m,&n)&&m) {

        for(int i=1;i<=m;i++) scanf("%s",s[i]+1);

        parse_st_ed();

        int ans=bfs();

        if(ans==-1) puts("Impossible");

        else printf("%d\n",ans);

    }

    return 0;

}

巴特西

Bloxorz I (poj3322) (BFS)

【题目描述】

【算法】

【题目链接】

【代码】

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