UVa216

来源：互联网发布：淘宝客服发展前景编辑：程序博客网时间：2024/05/23 14:51

纯粹枚举

枚举所有排列，记录最小的距离。

不过貌似有单源最短路径算法。

#include<stdio.h>#include<math.h>#define Maxnode 9#define INF 100000int ans[Maxnode];int a[Maxnode];double distance[Maxnode],ansdistance[Maxnode];int nodes;double maxdist;struct node{int x,y;}vector[Maxnode];void dfs(int cur){int i,ok,j;double xdis,ydis,dist,total_dis;if(cur==nodes+1){for(i=1,total_dis=0;i<nodes;i++){xdis=vector[a[i]].x-vector[a[i+1]].x;ydis=vector[a[i]].y-vector[a[i+1]].y;dist=sqrt(xdis*xdis+ydis*ydis);distance[i]=dist=dist+16;total_dis +=dist;}if(total_dis<maxdist){maxdist=total_dis;for(i=1;i<=nodes;i++){ans[i]=a[i];}for(i=1;i<nodes;i++){ansdistance[i]=distance[i];}}return ;}else{                               //枚举所有的排列for(i=1;i<=nodes;i++){ok=1;a[cur]=i;for(j=1;j<cur;j++){if(a[j]==a[cur]){ok=0;break;}}if(ok)dfs(cur+1);}}}int main(){int i,case_num=1;while(scanf("%d",&nodes)!=EOF&&nodes){maxdist=INF;for(i=1;i<=nodes;i++){scanf("%d %d",&vector[i].x,&vector[i].y);}dfs(1);printf("**********************************************************\n");printf("Network #%d\n",case_num);for(i=1;i<nodes;i++){printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2f feet.\n",vector[ans[i]].x,vector[ans[i]].y,vector[ans[i+1]].x,vector[ans[i+1]].y,ansdistance[i]);}printf("Number of feet of cable required is %.2f.\n",maxdist);case_num++;}return 0;}