bzoj4394【Usaco2015 Dec】Bessie

bzoj4394【Usaco2015 Dec】Bessie

Description

After eating too much fruit in Farmer John's kitchen, Bessie the cow is getting some very strange dreams! In her most recent dream, she is trapped in a maze in the shape of an N×M grid of tiles (1≤N,M≤1,000). She starts on the top-left tile and wants to get to the bottom-right tile. When she is standing on a tile, she can potentially move to the adjacent tiles in any of the four cardinal directions.

But wait! Each tile has a color, and each color has a different property! Bessie's head hurts just thinking about it:

If a tile is red, then it is impassable.
If a tile is pink, then it can be walked on normally.
If a tile is orange, then it can be walked on normally, but will make Bessie smell like oranges.
If a tile is blue, then it contains piranhas that will only let Bessie pass if she smells like oranges.
If a tile is purple, then Bessie will slide to the next tile in that direction (unless she is unable to cross it). If this tile is also a purple tile, then Bessie will continue to slide until she lands on a non-purple tile or hits an impassable tile. Sliding through a tile counts as a move. Purple tiles will also remove Bessie's smell.

(If you're confused about purple tiles, the example will illustrate their use.)

Please help Bessie get from the top-left to the bottom-right in as few moves as possible.

奶牛Bessie被困在了N*M的網格圖迷宮中，她位於左上角(1,1)，出口在右下角(N,M)。Bessie只能上下左右行走。

每塊地磚都有一個顏色：

如果是紅色，那麼不可通行。

如果是粉色，那麼可以通行。

如果是橙色，那麼可以通行，但是會給Bessie帶上橘子的氣味。

如果是藍色，那麼當且僅當Bessie帶著橘子的氣味時，才可以通行。

如果是紫色，那麼Bessie會保持原有方向滑過去，如果之後仍然是紫色，那麼會繼續滑。當滑到不是紫色的地磚上或者不可通行的時候，才會停下來。並且這會消除Bessie身上的氣味。每一步滑行和走路一樣，都需要耗費一單位時間。

請輸出Bessie逃到出口所需的最短時間。

Input

The first line has two integers N and M, representing the number of rows and columns of the maze.

The next NN lines have MM integers each, representing the maze:

The integer '0' is a red tile
The integer '1' is a pink tile
The integer '2' is an orange tile
The integer '3' is a blue tile
The integer '4' is a purple tile

The top-left and bottom-right integers will always be '1'.

Output

A single integer, representing the minimum number of moves Bessie must use to cross the maze, or -1 if it is impossible to do so.

Sample Input

Sample Output

HINT

In this example, Bessie walks one square down and two squares to the right (and then slides one more square to the right). She walks one square up, one square left, and one square down (sliding two more squares down) and finishes by walking one more square right. This is a total of 10 moves (DRRRULDDDR).

Source

這是標程

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

int dr[] = {-1, 0, 1, 0};
int dc[] = {0, -1, 0, 1};

struct state {
  int r;
  int c;
  int ld;
  bool smell;

  state(int r, int c, int ld, bool smell) :
      r(r), c(c), ld(ld), smell(smell) {
  }

  int pack() {
    return (smell ? 1 : 0) + 2 * (ld + 1) + 10 * c + 10000 * r;
  }

  static state unpack(int x) {
    return state(x / 10000, (x / 10) % 1000,
                 (x / 2) % 5 - 1, x & 1);
  }
};

int getcell(const vector >& A, int r, int c) {
  if (r < 0 || r >= A.size() || c < 0 || c >= A[r].size()) {
    return 0;
  }
  return A[r][c];
}

int main() {
  ios_base::sync_with_stdio(false);

  int N, M;
  cin >> N >> M;
  vector > A(N, vector(M));
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      cin >> A[i][j];
    }
  }
  queue q;
  vector D(10000000, -1);

  int s = state(0, 0, -1, false).pack();
  q.push(s);
  D[s] = 0;

  while (!q.empty()) {
    state st = state::unpack(q.front());
    q.pop();
//    cout<
 
這是我的程序，有兩個點WA...不知道為什麼
 
#include
#include
#include
#include
#include
#include
#include
#define F(i,j,n) for(int i=j;i<=n;i++)
#define maxn 1100
#define maxm 20000100
using namespace std;
int n,m,a[maxn][maxn],d[maxm];
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};
struct node
{
	int x,y,dir,sme;
}now,nxt;
queue q;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline int num(node tmp)
{
	return tmp.sme+2*(tmp.dir+1)+10*(tmp.x-1)+10000*(tmp.y-1);
}
int main()
{
	n=read();m=read();
	F(i,1,n) F(j,1,m) a[i][j]=read();
	memset(d,-1,sizeof(d));
	now=(node){1,1,-1,0};
	d[num(now)]=0;
	q.push(now);
	while (!q.empty())
	{
		now=q.front();q.pop();
		cout<n||ty<1||ty>m) continue;
			if (a[tx][ty]!=0&&a[tx][ty]!=3)
			{
				nxt=(node){tx,ty,now.dir,a[tx][ty]==2};
				if (d[num(nxt)]!=-1) continue;
				d[num(nxt)]=d[num(now)]+1;
				q.push(nxt);
				continue;
			}
		}
		F(i,0,3)
		{
			int tx=now.x+dx[i],ty=now.y+dy[i];
			if (tx<1||tx>n||ty<1||ty>m) continue;
			if (a[tx][ty]==0||(a[tx][ty]==3&&!now.sme)) continue;
			int ts=a[tx][ty]==2?1:(a[tx][ty]==4?0:now.sme);
			nxt=(node){tx,ty,i,ts};
			if (d[num(nxt)]!=-1) continue;
			d[num(nxt)]=d[num(now)]+1;
			q.push(nxt);
		}
	}
	printf("-1\n");
	return 0;
}