HDU 3137 No Left Turns（BFS）

标签：hdu   bfs   

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3137

ALL HEADS: You‘re a Knight of the Round Table?

ROBIN: I am.

LEFT HEAD: In that case I shall have to kill you.

MIDDLE HEAD: Shall I?

RIGHT HEAD: Oh, I don‘t think so.

MIDDLE HEAD: Well, what do I think?

LEFT HEAD: I think kill him.

RIGHT HEAD: Well let‘s be nice to him.

MIDDLE HEAD: Oh shut up.

As the story goes, the Knight scarpers off. Right Head has taken it upon himself to search the grounds for the knight so he, Left, and Middle can go extinguish him (and then have tea and biscuits.)

Consider the following 8 by 12 maze, where shaded squares are walls that can’t be entered.


The shortest path between the Right Head (denoted by the S, for start) and the knight (denoted by the F, for finish) is of length 3, as illustrated above. But! Right Head can’t turn left or make UTurns. He can only move forward and turn right. That means the shortest path that Right Head can find is significantly longer: at 29!

XXXXXXXXXXXXXX
X          XXX
X XFXXXXX    X
XXX   XX  XX X
X S          X
XX  XXXXXX X X
X        X X X
X X      X X X
XXX XX       X
XXXXXXXXXXXXXX

1
10 14
XXXXXXXXXXXXXX
X          XXX
X XFXXXXX    X
XXX   XX  XX X
X S          X
XX  XXXXXX X X
X        X X X
X X      X X X
XXX XX       X
XXXXXXXXXXXXXX

29

题意：

最少的步数从S走到F，每次只能走当前方向的右方或者直走！

X代表不能走！

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn = 26;
int xx[4] = {-1,0,1,0};
int yy[4] = {0,1,0,-1};
int vis[5][maxn][maxn];
char mm[maxn][maxn];
int n, m;
int x1, y1;//起点
int x2, y2;//终点
struct node
{
    int x;
    int y;
    int step;
    int dir;//标记方向
};
int judge(int a, int b)
{
    if(a>=0 && a<n && b>=0 && b<m)
        return 1;
    return 0;
}
int BFS()
{
    memset(vis,-1,sizeof(vis));
    node frontt, next;
    queue<node>p,q;
    int tx, ty;
    for(int i = 0; i < 4; i++)
    {
        tx = x1 + xx[i];
        ty = y1 + yy[i];
        if(mm[tx][ty] != 'X')
        {
            frontt.x = tx;
            frontt.y = ty;
            frontt.step = 0;
            frontt.dir = i;
            q.push(frontt);
            vis[i][tx][ty] = 1;
        }
    }
    while(!q.empty())
    {
        frontt = q.front();
        q.pop();
        if(frontt.x == x2 && frontt.y == y2)
        {
            return frontt.step+1;
        }
        for(int i = 1; i <= 2; i++)
        {
            if(i == 1)
            {
                next.dir = frontt.dir;//当前方向继续向前
            }
            else
            {
                next.dir = (frontt.dir+1)%4;//向当前方向的右方
            }
            tx = frontt.x + xx[next.dir];
            ty = frontt.y + yy[next.dir];
            next.step = frontt.step+1;
            if(mm[tx][ty]!='X' && judge(tx,ty) && (next.step<vis[next.dir][tx][ty] || vis[next.dir][tx][ty]==-1))
            {
                next.x = tx;
                next.y = ty;
                vis[next.dir][tx][ty] = next.step;
                q.push(next);
            }
        }
    }
}
int main()
{
#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        getchar();
        for(int i = 0; i < n; i++)
        {
            gets(mm[i]);
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(mm[i][j] == 'S')
                {
                    x1 = i;
                    y1 = j;
                }
                else if(mm[i][j] == 'F')
                {
                    x2 = i;
                    y2 = j;
                }
            }
        }
        int ans = BFS();
        printf("%d\n",ans);
    }
    return 0;
}

HDU 3137 No Left Turns（BFS）

标签：hdu   bfs   

原文地址：http://blog.csdn.net/u012860063/article/details/40833649