点集配对问题 状压DP

大白P61，空间里有n个点P0，P1，...，Pn-1，把它们两两配对，要求所有的点对距离之和最小。其中n<=20.

分析：dp[s]表示集合s配对后的最小距离之和，状态转移方程为dp[s]=min(dp[s],dp[s^(1<<i)^(1<<j)]+dis(i,j))，其中i<j。

注意这里i是不需要枚举的，因为i最后都是要配对的，无须枚举。

代码：

#include <iostream>#include <functional>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i--)#define FRL(i,a,b)  for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")typedef long long ll;using namespace std;#define INF 1<<30#define mod 1000000009const int maxn = 1005;const int MAXN = 2005;const int MAXM = 200010;const int N = 1005;struct Node{    double x,y;}node[22];int n;double dp[(1<<20)+10];double Dis(Node a,Node b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}void solve(){    int i,j;    int m=1<<n;    for (i=0;i<m;i++) dp[i]=INF;    dp[0]=0;    for (int s=0;s<m;s++)    {        for (i=0;i<n;i++)            if (s&(1<<i)) break;        for (j=i+1;j<n;j++)        {            if (s&(1<<j))            {                dp[s]=min(dp[s],dp[s^(1<<i)^(1<<j)]+Dis(node[i],node[j]));            }        }    }}int main(){//#ifndef ONLINE_JUDGE//    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);//#endif    int i,j;    while (~scanf("%d",&n))    {        for (i=0;i<n;i++)            scanf("%lf%lf",&node[i].x,&node[i].y);        solve();        printf("%f\n",dp[(1<<n)-1]);    }    return 0;}