點集配對問題

標籤：dp

點集配對問題  空間裡n個點，使它們配成n/2對點，使得每個點恰好在一個點對中。

要求所有點隊中，兩點距離之和盡量下  n <= 20

d(s) = min(d{S - {i} - {j}+ |Pi Pj|  | j屬於S, j > i, i = min{S}}

//#pragma comment(linker, "/STACK:102400000,102400000")//HEAD#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <algorithm>#include <queue>#include <string>#include <set>#include <stack>#include <map>#include <cmath>#include <cstdlib>using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FED(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))//STL#define PB push_back//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)typedef long long LL;const int INF = 0x3f3f3f3f;const int MAXN = 1010;#define FF(i, a, b) for(int i = (a); i < (b); ++i)#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)#define ALL(c) (c).begin(), (c).end()#define SZ(V) (int)V.size()#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)#define sqr(x) (x) * (x)typedef vector <int> VI;typedef unsigned long long ULL;const double eps = 1e-10;const LL MOD = 1e9 + 7;int inx[25], iny[25], inz[25];double dp[1 << 20];int n;double dis(int x, int y){    return sqrt(sqr(inx[x] - inx[y]) + sqr(iny[x] - iny[y]) + sqr(inz[x] - inz[y]));}double dfs(int x){    if (dp[x] < INF)    return dp[x];    int i, u = -1, v = -1;    for (i = 0; i < n; i++)        if ((x & (1 << i)) == 0)            { u = i;   break;}    for (int j = i + 1; j < n; j++)    {        if ((x & (1 << j)) == 0)            dp[x] = min(dfs(x | (1 << j) | (1 << u)) + dis(u, j), dp[x]);    }    return dp[x];}int main(){    while (~RI(n))    {        int x, y, z;        REP(i, n)            RIII(inx[i], iny[i], inz[i]);        REP(i, 1 << n) dp[i] = INF;        dp[(1 << n) - 1] = 0;        dfs(0);        printf("%.10lf\n", dp[0]);    }    return 0;}