joj2558

Result TIME Limit MEMORY Limit Run Times AC Times JUDGE

8s 65536K 395 55 Standard


For a boat on a large body of water, strong currents can be dangerous, but with careful planning, they can be harnessed to help the boat reach its destination. Your job is to help in that planning.


At each location, the current flows in some direction. The captain can choose to either go with the flow of the current, using no energy, or to move one square in any other direction, at the cost of one energy unit. The boat always moves in one of the following eight directions: north, south, east, west, northeast, northwest, southeast, southwest. The boat cannot leave the boundary of the lake. You are to help him devise a strategy to reach the destination with the minimum energy consumption.


Input Specification


The input contains several test cases. Each case starts with two integers r and c(The lake is represented as a rectangular grid), the number of rows and columns in the grid. The grid has no more than 1000 rows and no more than 1000 columns. Each of the following r lines contains exactly c characters, each a digit from 0 to 7 inclusive. The character 0 means the current flows north (i.e. up in the grid, in the direction of decreasing row number), 1 means it flows northeast, 2 means east (i.e. in the direction of increasing column number), 3 means southeast, and so on in a clockwise manner:
7 0 1
 \|/
6-*-2
 /|\
5 4 3
The line after the grid contains a single integer n, the number of trips to be made, which is at most 50. Each of the following n lines describes a trip using four integers rs, cs, rd, cd, giving the row and column of the starting point and the destination of the trip. Rows and columns are numbered starting with 1.
Sample Input


5 5
04125
03355
64734
72377
02062
3
4 2 4 2
4 5 1 4
5 3 3 4
Output Specification


For each trip in a test case, output a line containing a single integer, the minimum number of energy units needed to get from the starting point to the destination. The output of each test case is ended with an empty line.
Output for Sample Input


0
2
1


Problem Source: Waterloo


This problem is used for contest: 129  149 
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Problem Set with Online Judge System Version 3.12
Jilin University


Developed by skywind, SIYEE

joj2558是至今为止让我最蛋疼的一个题，我做的3个晚上才把他搞定但是之后心情暴爽。我感觉ACM这点是最吸引我的。。在做这个题的途中我也明白了多两个变量

难度增加一倍。好了不扯淡了。说正经的了。

这是一个宽度优先搜索的问题，一开始我还认为是用优先队列做。做了一个晚上没有做出来后来意识到队列与bfs的组合的实质才想出来的。之所以用队列与bfs结合我

现在真正的理解了是通过队列将所有元素按照权重依次排成一列。所以这道题用优先队列是不行的。因为当一个点入队之后还可能跟他有相同权重的还要进去。然而

优先队列不能保证这点能够实现。。因为优先队列在一个点入队之后得向上下左右四个方向扩展然而这四个方向只有一个是跟这个点有相同权重的其他的都要比这个点大。

#include<cstdio>

#include<cstring>
#include<queue>
using namespace std;
int m,n;
int map[1020][1020];
int visited[1020][1020];
class Node
{
public:
int x,y;
int step;
};
int move[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
void  bfs(Node node1,Node node2)
{
queue<Node>q;
while(!q.empty())q.pop();
node1.step=0;
visited[node1.x][node1.y]=1;
q.push(node1);
while(!q.empty())
{
Node temp;
Node temp1;
temp=q.front();
q.pop();//printf("%d  %d   %d \n",temp.x,temp.y,temp.step);
if(temp.x==node2.x&&temp.y==node2.y)
{
printf("%d\n",temp.step);return ;
}
int xx=temp.x;
int yy=temp.y;
int step=temp.step;
while(1)
{
int x=xx+move[map[xx][yy]][0];
int y=yy+move[map[xx][yy]][1];
if(x>0 &&x<=m&&y>0&&y<=n
&&!visited[x][y])
{
visited[x][y]=1;
temp1.x=x;temp1.y=y;temp1.step=step;
q.push(temp1);
if(x==node2.x&&y==node2.y)
{
printf("%d\n",step);
return ;
}
xx=x;yy=y;
}
else break;
}
for(int i=0;i<8;i++)
{
int x=temp.x+move[i][0];
int y=temp.y+move[i][1];
if(x>0 &&x<=m&&y>0&&y<=n
&&!visited[x][y])
{
visited[x][y]=1;
temp1.x=x;
temp1.y=y;
temp1.step=temp.step+1;
q.push(temp1);
if(x==node2.x&&y==node2.y)
{
printf("%d\n",temp1.step);return ;
}
while(1)
{
xx=x+move[map[x][y]][0];
yy=y+move[map[x][y]][1];
if(xx>0&&xx<=m&&yy>0&&yy<=n&&!visited[xx][yy])
{
visited[xx][yy]=1;
temp1.x=xx;
temp1.y=yy;
temp1.step=temp.step+1;
q.push(temp1);
x=xx;y=yy;
if(xx==node2.x&&yy==node2.y)
{
printf("%d\n",temp1.step);return ;
}
}
else break;
}
}
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
char str[1020];
while(scanf("%d%d",&m,&n)==2)
{
getchar();
for(int i=1;i<=m;i++)
{
gets(str);
for(int j=1;j<=n;j++)
{
map[i][j]=str[j-1]-'0';
}
}
int num;
scanf("%d",&num);
Node node1,node2;
for(int i=0;i<num;i++)
{
memset(visited,0,sizeof(visited));
scanf("%d%d%d%d",&node1.x,&node1.y,&node2.x,&node2.y);
bfs(node1,node2);
}
printf("\n");
}
return 0;
}