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最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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文章标签： java

原文链接：http://www.cnblogs.com/stepping/p/5679290.html

本文介绍了一个关于懒惰狗狗寻找从家到朋友家最短路径的问题。通过使用优先队列实现的广度优先搜索算法，文章详细解释了如何确定狗狗需要挖掘的最少土地格子数。

Dogs

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 8 Accepted Submission(s) : 2

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Problem Description

Prairie dog comes again! Someday one little prairie dog Tim wants to visit one of his friends on the farmland, but he is as lazy as his friend (who required Tim to come to his place instead of going to Tim's), So he turn to you for help to point out how could him dig as less as he could.

We know the farmland is divided into a grid, and some of the lattices form houses, where many little dogs live in. If the lattices connect to each other in any case, they belong to the same house. Then the little Tim start from his home located at (x0, y0) aim at his friend's home ( x1, y1 ). During the journey, he must walk through a lot of lattices, when in a house he can just walk through without digging, but he must dig some distance to reach another house. The farmland will be as big as 1000 * 1000, and the up left corner is labeled as ( 1, 1 ).

Input

The input is divided into blocks. The first line in each block contains two integers: the length m of the farmland, the width n of the farmland (m, n ≤ 1000). The next lines contain m rows and each row have n letters, with 'X' stands for the lattices of house, and '.' stands for the empty land. The following two lines is the start and end places' coordinates, we guarantee that they are located at 'X'. There will be a blank line between every test case. The block where both two numbers in the first line are equal to zero denotes the end of the input.

Output

For each case you should just output a line which contains only one integer, which is the number of minimal lattices Tim must dig.

Sample Input

6 6
..X...
XXX.X.
....X.
X.....
X.....
X.X...
3 5
6 3

0 0

Sample Output

Hint

Hint: Three lattices Tim should dig: ( 2, 4 ), ( 3, 1 ), ( 6, 2 ).

Source

2009 Multi-University Training Contest 1 - Host by TJU

思路: 采用优先队列，哪个点时间少哪个点先处理

#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m,i,j,sx,sy,tx,ty;
char mp[1005][1005];
bool vis[1005][1005];  //int爆内存，用bool型

struct node
{
    int x,y,ti;
    node(int a,int b,int c){x=a;y=b;ti=c;}
};
int dr[4][2]={{1,0},{-1,0},{0,1},{0,-1} };
struct cmp
{
    bool operator()(node a,node b)
    {
        return a.ti>b.ti;
    }
};
bool check(int x,int y)
{
    if(x>=0 && x<n && y>=0 && y<m && !vis[x][y])  return 1;
     return 0;
}
int bfs()
{
    priority_queue<node,vector<node>,cmp> Q;
    Q.push(node(sx,sy,0));
    vis[sx][sy]=1;
    while(!Q.empty())
    {
        node p=Q.top();
        Q.pop();
       for(int i=0;i<4;i++)
       {
           int xx=p.x+dr[i][0];
           int yy=p.y+dr[i][1];
           if(!check(xx,yy)) continue;
           vis[xx][yy]=1;
           int ti=0;
           if (mp[xx][yy]=='X') ti=p.ti;
                  else ti=p.ti+1;
             Q.push(node(xx,yy,ti));
           if (xx==tx && yy==ty) return ti;
       }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m))
    {
        if(n==0 && m==0) break;//以0 0 结束
        for(i=0;i<n;i++)
         scanf("%s",&mp[i]);
            scanf("%d%d",&sx,&sy);
            scanf("%d%d",&tx,&ty);
            sx--;
            sy--;
            tx--;
            ty--;
            memset(vis,0,sizeof(vis));
            if (sx==tx && sy==ty) printf("0\n");
                        else  printf("%d\n",bfs());
    }

    return 0;
}