HDU3362+状态压缩

dp[ i ]表示该状态下得所需花费。

/*
状态压缩dp
dp[i] = min( dp[ i-j ]+cost[ j ] );
由i-j的状态转到i的状态
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<math.h>
using namespace std;
const int maxn = 20;
const double inf = 99999999.0;
const int maxm = (1<<20);
const double eps = 1e-8;
struct node{
    double x,y;
}dot[ maxn ];
double dp[ maxm ];
double dis[ maxn ][ maxn ];
double dist( int i,int j ){
    return sqrt( ( dot[i].x-dot[j].x )*( dot[i].x-dot[j].x )+( dot[i].y-dot[j].y )*( dot[i].y-dot[j].y ) );
}
int main(){
    int n;
    while( scanf("%d",&n),n ){
        int sum,cnt,flag;
        cnt = 0;
        sum = 0;
        for( int i=0;i<n;i++ ){
            scanf("%lf%lf%d",&dot[i].x,&dot[i].y,&flag);
            if( flag==1 ){
                sum += (1<<i);
                cnt++;
            }
        }
        if(( n>1&&cnt<2 )||( n==1&&cnt==0 )){
            printf("No Solution\n");
            continue;
        }
        for( int i=0;i<n;i++ )
            for( int j=0;j<n;j++ )
                dis[ i ][ j ] = dist( i,j );
        int N = (1<<n);
        for( int i=0;i<N;i++ )
            dp[ i ] = inf;
        dp[ sum ] = 0.0;
        for( int i=sum;i<N;i++ ){
            if( dp[i]>inf ) continue;
            for( int j=0;j<n;j++ ){
                if( i&(1<<j) )//j是固定的
                    continue;
                double min1 = inf;
                double min2 = inf;
                for( int k=0;k<n;k++ ){//找出min1,min2
                    if( i&(1<<k) ){
                        if( min1>dis[ j ][ k ] ){
                            min2 = min1;
                            min1 = dis[ j ][ k ];
                        }
                        else if( min2>dis[ j ][ k ] ){
                            min2 = dis[ j ][ k ];
                        }
                    }
                }
                dp[ i|(1<<j) ] = min( dp[ i|(1<<j) ],dp[i]+min1+min2 );
            }
        }
        printf("%.6lf\n",dp[ N-1 ]);
    }
    return 0;
}