UVA 216 Getting in Line

UVA_216

    这个题目可以直接深搜，深搜的边界就是加入line的点为n，当总长度比标记过的长度小的时候，即可更新记录line的数组和总长度。

#include<stdio.h>
#include<string.h>
#include<math.h>
int n,x[10],y[10],vis[10];
int temp[10],ans[10];
double total;
void dfs(int cur,double t)
{
    int i;
    double k,t_x,t_y;
    if(cur==n)
    {
        if(t<total)
        {
            total=t;
            memcpy(ans,temp,sizeof(temp));
        }
        return;    
    }
    for(i=0;i<n;i++)
        if(!vis[i])
        {
            vis[i]=1;
            temp[cur]=i;
            t_x=x[temp[cur]]-x[temp[cur-1]];
            t_y=y[temp[cur]]-y[temp[cur-1]];
            dfs(cur+1,t+sqrt(t_x*t_x+t_y*t_y));
            vis[i]=0;
        }
}
int main()
{
    int i,j,k,t,t_x,t_y;
    t=0;
    while(1)
    {
        scanf("%d",&n);
        if(n==0)
            break;
        for(i=0;i<n;i++)
            scanf("%d%d",&x[i],&y[i]);
        memset(vis,0,sizeof(vis));
        total=1000000000.0;
        for(i=0;i<n;i++)
        {
            vis[i]=1;
            temp[0]=i;
            dfs(1,0.0);
            vis[i]=0;
        }
        printf("**********************************************************\n");
        printf("Network #%d\n",++t);
        for(i=0;i<n-1;i++)
        {
            t_x=x[ans[i+1]]-x[ans[i]];
            t_y=y[ans[i+1]]-y[ans[i]];
            printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2f feet.\n",
            x[ans[i]],y[ans[i]],x[ans[i+1]],y[ans[i+1]],16+sqrt(t_x*t_x+t_y*t_y));
        }
        printf("Number of feet of cable required is %.2f.\n",total+16*(n-1));
    }
    return 0;    
}