经典状压DP
设 f [ j ] [ i ] f[j][i] f[j][i] 表示在第 j j j 个点的状态为 i i i,
可得动态转移方程为:
f [ j ] [ i ] = min ( f [ j ] [ i ] , f [ k ] [ i − ( 1 < < j − 1 ) ] + d i s ( j , k ) ) f[j][i]=\min(f[j][i],f[k][i-(1<<j-1)]+dis(j,k)) f[j][i]=min(f[j][i],f[k][i−(1<<j−1)]+dis(j,k))
代码
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
double x[201],y[201],f[20][1<<15];
int n;
double dis(int a,int b)
{
return sqrt(((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b])));
}
int main()
{
cin>>n;
for(int i=1; i<=n; i++)
scanf("%lf%lf",&x[i],&y[i]);
for(int i=1; i<=n; i++)
for(int j=1; j<=(1<<n)-1; j++) //求最小,初始化最大
f[i][j]=2147483647;
for(int i=0; i<=(1<<n)-1; i++)
for(int j=1; j<=n; j++)
{
if(!(i&(1<<j-1))) //这个点为空
continue;
if(i==(1<<j-1)) //已经走过了
{
f[j][i]=0;
continue;
}
for(int k=1; k<=n; k++)
{
if(j==k||!(i&(1<<k-1))) //已经走过此点或此点为空
continue;
f[j][i]=min(f[j][i],f[k][i-(1<<j-1)]+dis(k,j));
}
}
double ans=101010101;
for(int i=1; i<=n; i++)
ans=min(ans,f[i][(1<<n)-1]+dis(i,0)); //加上(0,0)起点
printf("%.2lf",ans);
return 0;
}