Central Avenue Road+csuoj+水题

Central Avenue Road

Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 30  Solved: 10
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Description

There are a lot of trees in the park of Jilin University. We want to open up a road in this forest in a straight line from one tree to another. In order to view more scenery, the difference of both sides of the road is minimum (zero or one). If other tree is just on the road, it is not counted on both sides. Of course, we have a lot of ways to build this road. The best way is that the distance of two reference trees is shortest.

Input

The first line of each case is the number of trees N (2 <= N <=100). The next N lines have two integers that is the x and y coordinates of the i-th tree. The input file is terminated by N=0.

Output

For each case, output the distance of the two reference trees that the road obeys the above rules and the distance is shortest. Keep three digits after decimal point.

Sample Input

40 03 410 00 100

Sample Output

5.000

解决方案：求建一条两颗树之间的路，要求左边的树的数目和右边的数目之差最多不能超过1。暴力即可，但要特判斜率为0和为无穷的情况。

code：
#include <iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;const int maxn=105;int x[maxn],y[maxn];int n;double dist(int i,int j){    return  sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])+0.000001);}bool judge(int i,int j){    int a=y[j]-y[i];    int b=x[i]-x[j];    int c=x[j]*y[i]-x[i]*y[j];    int u=x[i]-x[j];    int d=y[i]-y[j];    int lf=0,rt=0;    if(u==0&&d!=0)    {        for(int k=0; k<n; k++)        {            if(x[k]<x[i])            {                lf++;            }            if(x[k]>x[i])            {                rt++;            }        }    }    else if(d==0&&u!=0)    {        for(int k=0; k<n; k++)        {            if(y[k]<y[i])            {                lf++;            }            if(y[k]>y[i])            {                rt++;            }        }    }    else    {        for(int k=0; k<n; k++)        {            if(a*x[k]+b*y[k]+c>0)            {                lf++;            }            else if(a*x[k]+b*y[k]+c<0)            {                rt++;            }        }    }    if(fabs(rt-lf)<=1) return true;    else return false;}int main(){    while(~scanf("%d",&n)&&n)    {        for(int i=0; i<n; i++)        {            scanf("%d%d",&x[i],&y[i]);        }        double Max=0x3f3f3f3f3f;        for(int i=0;i<n;i++){            for(int j=i+1;j<n;j++){                if(judge(i,j)){                    double temp=dist(i,j);                    if(temp<Max)                        Max=temp;                }            }        }        printf("%.3lf\n",Max);    }    return 0;}