UVa 216 Getting in Line
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入门经典里推荐的有关暴力求解的题目,正好新学了STL里有一个算法生成下一个排列。
题目里点最多8个,排列量最大8!,只有4万多,暴力求解可行,第一次在UVa上做题,,努力。
#include<cstdio>#include <cmath>#include <algorithm>using namespace std;struct Points{ double x,y,dis;}a[10],res[10];//a用来输入,res保存结果double getdis(Points &a, Points &b)//计算距离{ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))+16;}int main(){ int n,k=1; while(~scanf("%d",&n) && n) { printf("**********************************************************\n"); printf("Network #%d\n",k++); int i,data[10]; double Min=2147483645,resd[10]; for(i=0;i<n;i++) { scanf("%lf%lf",&a[i].x,&a[i].y); data[i]=i; //记录每个点的索引 } do{ double sum=0,tmpdis[10]; for(i=0;i<n-1;i++) { tmpdis[i]=getdis(a[data[i]],a[data[i+1]]); sum+=tmpdis[i]; } if(sum<Min) // 如果找到更小的距离,保存点坐标与点间距 { for(int j=0;j<n-1;j++) { resd[j]=tmpdis[j]; res[j]=a[data[j]]; } res[n-1]=a[data[n-1]]; Min=sum; } }while (next_permutation(data,data+n));//下一个排列 for(i=0;i<n-1;i++) printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n",res[i].x,res[i].y,res[i+1].x,res[i+1].y,resd[i]); printf("Number of feet of cable required is %.2lf.\n",Min); } return 0;}
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