poj 3626 Mud Puddles (简单BFS)

Mud Puddles

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2743 Accepted: 1607

Description

Farmer John is leaving his house promptly at 6 AM for his daily milking of Bessie. However, the previous evening saw a heavy rain, and the fields are quite muddy. FJ starts at the point (0, 0) in the coordinate plane and heads toward Bessie who is located at (X, Y) (-500 ≤ X ≤ 500; -500 ≤ Y ≤ 500). He can see all N (1 ≤ N ≤ 10,000) puddles of mud, located at points (A_i, B_i) (-500 ≤ A_i ≤ 500; -500 ≤ B_i ≤ 500) on the field. Each puddle occupies only the point it is on.

Having just bought new boots, Farmer John absolutely does not want to dirty them by stepping in a puddle, but he also wants to get to Bessie as quickly as possible. He's already late because he had to count all the puddles. If Farmer John can only travel parallel to the axes and turn at points with integer coordinates, what is the shortest distance he must travel to reach Bessie and keep his boots clean? There will always be a path without mud that Farmer John can take to reach Bessie.

Input

* Line 1: Three space-separate integers: X, Y, and N.
* Lines 2..N+1: Line i+1 contains two space-separated integers: A_i and B_i

Output

* Line 1: The minimum distance that Farmer John has to travel to reach Bessie without stepping in mud.

Sample Input

1 2 70 2-1 33 11 14 2-1 12 2

Sample Output

11

Source

USACO 2007 December Silver

 

AC掉了的代码：

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

int x,y,n,vs[1001][1001];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
struct Node{
    int x;
    int y;
    int counter;
};

void bfs()
{
    int i;//用于选择4个方向;
    queue<Node>Q;//创建一个队列;
    Node p,q;
    p.x=500;
    p.y=500;
    p.counter=0;
    vs[p.x][p.y]=1;
    Q.push(p);//入队;
    while(!Q.empty())//当队列不为空的时候循环;
    {
        p=Q.front();
        Q.pop();
        for(i=0; i<4; i++)
        {
            q.x=p.x+dx[i];
            q.y=p.y+dy[i];
            q.counter=p.counter+1;
            if(q.x>=0 && q.x<=1000 && q.y>=0  && q.y<=1000 && vs[q.x][q.y]==0)
            {
                if(q.x==x && q.y==y)
                {
                    cout<<q.counter<<endl;
                    return;
                }
                vs[q.x][q.y]=1;
                Q.push(q);
            }
        }
    }
}

int main()
{
    int i,c,d;
    cin>>x>>y>>n;
    memset(vs, 0, sizeof(vs));
    x+=500;
    y+=500;
    for(i=0; i<n; i++)
    {
        cin>>c>>d;
        vs[c+500][d+500]=1;
    }
    if(x==500 && y==500)
        cout<<0<<endl;
    else
        bfs();
    return 0;
}