USACO 2012 March Silver Tractor /// 优先队列BFS oj21567

题目大意：

输入n，（x，y）；n为阻挡的草堆数量，（x，y）为开始时拖拉机所在的位置

接下来n行每行一个坐标（a，b）；为各个草堆的坐标

输出拖拉机要回到原点（0,0）需要移动的草堆数量

Sample Input

7 6 3
6 2
5 2
4 3
2 1
7 3
5 4
6 4

Sample Output

1

Hint

INPUT DETAILS:

The tractor starts at (6,3).  There are 7 bales of hay, at positions (6,2), (5,2), (4,3), (2,1), (7,3), (5,4), and (6,4).

OUTPUT DETAILS:

Farmer John only needs to remove one bale of hay to free his tractor.

 

直接从拖拉机的位置往外广搜整个矩阵 遇到边界就结束  遇到边界前的路径中最少草堆的一条有几个

优先队列升序排序可直接取出当前草堆数量最少的一条路

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#include <queue>
#include <functional>
using namespace std;
struct NODE
{
    int sp,x,y;
    NODE(){};
    NODE(int spi,int xi,int yi):
        sp(spi),x(xi),y(yi){};
    bool operator<(const NODE& p)const{
        return sp>p.sp;  /// 升序排序这里应该是 大于号！！
    }
};
int mov[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int G[1005][1005];
int st,ed,n;
bool bound(int x,int y)
{
    return x==0||y==0||x==1001||y==1001;
}
int main()
{
    while(~scanf("%d",&n))
    {
        scanf("%d%d",&st,&ed);
        memset(G,0,sizeof(G));
        for(int i=1;i<=n;i++)
        {
            int a,b; scanf("%d%d",&a,&b);
            G[a][b]=1;
        }

        priority_queue < NODE > q;
        q.push(NODE(0,st,ed));
        G[st][ed]=-1;
        while(!q.empty())
        {
            NODE tmp=q.top(); q.pop();

            if(bound(tmp.x,tmp.y))
            {
                printf("%d
",tmp.sp);
                break;
            }//printf("%d %d %d
",tmp.x,tmp.y,tmp.sp);

            for(int i=0;i<4;i++)
            {
                int nowx=tmp.x+mov[i][0],
                    nowy=tmp.y+mov[i][1],
                    nowsp=tmp.sp;

                if(G[nowx][nowy]==-1) continue;

                if(G[nowx][nowy]) nowsp++;
                G[nowx][nowy]=-1;

                q.push(NODE(nowsp,nowx,nowy));
                //printf("%d %d %d
",nowx,nowy,nowsp);
            }
        }
    }

    return 0;
}