HDU-4643-GSM(DFS)

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4 4
0 2
1 3
1 0
2 0
1 2
1 1
2 2
2 1
4
1 2
1 3
1 4
3 4

0
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Hint
The train way from a to b will not cross the point with the same distance from more than 2 base stations. 
(For the distance d1 and d2, if fabs(d1-d2)<1e-7, we think d1 == d2). 
And every city exactly receive signal from just one base station.
 

思路：每次获取中点的最近的基站的id，id跟哪边不一样就往哪边递归查找。

#include <stdio.h>
#define INF 99999999

int n,m,ans;
bool vis[50];
double sx[50],sy[50],ex[50],ey[50];

int get(double x,double y)//获取离该点最近的基站
{
    double mn=INF;
    int id,i;

    for(i=0;i<m;i++)
    {
        if((x-ex[i])*(x-ex[i])+(y-ey[i])*(y-ey[i])<mn)
        {
            mn=(x-ex[i])*(x-ex[i])+(y-ey[i])*(y-ey[i]);
            id=i;
        }
    }

    return id;
}

void dfs(int l,int r,double lx,double ly,double rx,double ry)//递归查找，每次获取中点的最近的基站的id，id跟哪边不一样就往哪边找。
{
    if((lx-rx)*(lx-rx)+(ly-ry)*(ly-ry)<1e-14) return;

    double x=(lx+rx)/2.0;
    double y=(ly+ry)/2.0;

    int id=get(x,y);

    if(!vis[id])
    {
        vis[id]=1;
        ans++;
    }

    if(id!=l) dfs(l,id,lx,ly,x,y);

    if(id!=r) dfs(id,r,x,y,rx,ry);
}

int main()
{
    int i,q,id1,id2;
    int from,to;

    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<n;i++) scanf("%lf%lf",&sx[i],&sy[i]);

        for(i=0;i<m;i++) scanf("%lf%lf",&ex[i],&ey[i]);

        scanf("%d",&q);

        while(q--)
        {
            scanf("%d%d",&from,&to);

            id1=get(sx[from-1],sy[from-1]);
            id2=get(sx[to-1],sy[to-1]);

            if(id1==id2) printf("0\n");
            else
            {
                for(i=0;i<m;i++) vis[i]=0;
                vis[id1]=1;
                vis[id2]=1;
                ans=1;
                dfs(id1,id2,sx[from-1],sy[from-1],sx[to-1],sy[to-1]);
                printf("%d\n",ans);
            }
        }
    }
}

HDU-4643-GSM(DFS),布布扣,bubuko.com

HDU-4643-GSM(DFS)

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原文地址：http://blog.csdn.net/faithdmc/article/details/37669325