zoj 2097 Walking on a Chessboard

Walking on a Chessboard

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

A mini robot is put on a cell of a chessboard. The size of the chessboard is 8 * 8. The robot has four states denoted by integer from 1 to 4.

There is an integer in each cell of the chessboard, the integers are positive and not greater than 1000. The robot can walk up, down, left or right to one of the neighboring cells. The cost of the movement is the multiplication of the integer in the new cell and its original state, then the state of the robot is altered to (cost MOD 4 + 1). The initial state of the robot is 1.

Your task is to find the path between two given cells with the minimum total cost. Total cost is the sum of the costs for each walk along the path.

Input

The input contains multiple test cases. Each test case begins with a line containing the row number and column number of the start cell and the target cell. Row number and column number will be within 1 and 8. 8 lines follow, each with 8 integers. This is the chessboard configuration.

There will be a blank line between subsequent test cases. The last line of the input will have 4 zeros which signals the end of the input and should not be processed.

Output

For each test case, print the minimum total cost on a line.

Sample Input

1 1 2 2
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1

0 0 0 0

Sample Output

3

---

Author: CHEN, Shunbao

Source: Zhejiang University Local Contest 2004

这是一道关于BFS的题，与一般BFS的题不同的是，每走一步，它的状态都会改变，而题目要求的是从一个点到另一个点的最短距离

他每走一步的cost = 原来的状态 * 当前点的值

而他的状态state也更新为 = cost % 4 + 1

所以我们不能用纯BFS来做，我们需要记录到达每个点所花费的时间 

我用status[8][8][5] 来记录这个robot以某个状态到达某个点的最短距离

例如： status[temp.x-1][temp.y-1][temp.state] 就表示robot到达temp.x-1,temp.y-1，并且状态变为temp.state时，所花费的最短时间

下面是我的代码：(自我感觉写的不好)

#include<iostream>
#include<algorithm>
#include<memory.h>
#include<queue>
using namespace std;

struct point
{
	int x,y;
	int state;
	int val;
}start, temp, other;

int n1,n2,m1,m2;
int num[8][8];
int status[8][8][5];
int direction[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int cost;

void BFS()
{
	queue<point> que;
	start.x = n1;
	start.y = n2;
	start.state = 1;
	start.val = 0;
	que.push(start);

	while (!que.empty())
	{
		temp = que.front();
		que.pop();
		if( temp.x == m1 && temp.y == m2) 
		{
			if(temp.val < cost)
				cost = temp.val;
		}

		for(int i=0; i<4; i++)
		{
			int x = temp.x + direction[i][0];
			int y = temp.y + direction[i][1];
			if( x>0 && x<=8 && y>0 && y<=8 )
			{
				other.x = x;
				other.y = y;
				other.val = temp.val + temp.state * num[x-1][y-1];
				other.state = (temp.state * num[x-1][y-1]) % 4 + 1;	
				if( status[x-1][y-1][other.state] == 0 || other.val < status[x-1][y-1][other.state] )
				{
					status[x-1][y-1][other.state] = other.val;
					que.push(other);
				}
			}
		}
	}
}

int main()
{
	

	while( cin>>n1>>n2>>m1>>m2 && n1!=0 && n2!=0 && m1!=0 && m2!=0 )
	{
		for( int i=0; i<8; i++)
			for( int j=0; j<8; j++)
				cin>>num[i][j];	
		memset(status, 0, sizeof(status));
		cost = (1 << 30);
		BFS();
		cout<<cost<<endl;
	}
	return 0;
}