题目描述
房间里放着n块奶酪。一只小老鼠要把它们都吃掉，问至少要跑多少距离？老鼠一开始在(0,0)点处。
输入输出格式
输入格式：第一行一个数n (n<=15)
接下来每行2个实数，表示第i块奶酪的坐标。
两点之间的距离公式=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))
输出格式：一个数，表示要跑的最少距离，保留2位小数。
输入输出样例
输入样例#1：
复制
4 1 1 1 -1 -1 1 -1 -1
输出样例#1： 复制
7.41
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
struct node {
double x, y;
}nd[20];
double dp[17][(1 << 16) + 2];
double dis(int i, int j) {
return 1.0*sqrt((nd[i].x - nd[j].x)*(nd[i].x - nd[j].x) + (nd[i].y - nd[j].y)*(nd[i].y - nd[j].y));
}
int main()
{
// ios::sync_with_stdio(0);
n = rd(); ms(nd); mclr(dp, 127);
for (int i = 1; i <= n; i++) {
rdlf(nd[i].x); rdlf(nd[i].y);
}
for (int S = 0; S <= (1 << n) - 1; S++) {
for (int i = 1; i <= n; i++) {
if ((S&(1 << (i - 1))) == 0)continue;
if (S == (1 << (i - 1))) {
dp[i][S] = 0; continue;
}
for (int j = 1; j <= n; j++) {
if (i == j)continue;
if (S&(1 << (j - 1))) {
dp[i][S] = min(dp[i][S], 1.0*dis(i, j) + dp[j][S - (1 << (i - 1))]);
}
}
}
}
double MIN = 1.0*inf;
for (int i = 1; i <= n; i++) {
MIN = min(MIN, dis(0, i) + dp[i][(1 << n) - 1]);
}
printf("%.2lf
", 1.0*MIN);
return 0;
}