CF8C Looking for Order 题解

Posted 2020-11-10 zbsakioi

题目描述

Girl Lena likes it when everything is in order, and looks for order everywhere. Once she was getting ready for the University and noticed that the room was in a mess — all the objects from her handbag were thrown about the room. Of course, she wanted to put them back into her handbag. The problem is that the girl cannot carry more than two objects at a time, and cannot move the handbag. Also, if he has taken an object, she cannot put it anywhere except her handbag — her inherent sense of order does not let her do so.

You are given the coordinates of the handbag and the coordinates of the objects in some Сartesian coordinate system. It is known that the girl covers the distance between any two objects in the time equal to the squared length of the segment between the points of the objects. It is also known that initially the coordinates of the girl and the handbag are the same. You are asked to find such an order of actions, that the girl can put all the objects back into her handbag in a minimum time period.

输入格式

The first line of the input file contains the handbag‘s coordinates x_{s},y_{s}xs?,ys? . The second line contains number nn ( 1<=n<=241<=n<=24 ) — the amount of objects the girl has. The following nn lines contain the objects‘ coordinates. All the coordinates do not exceed 100 in absolute value. All the given positions are different. All the numbers are integer.

输出格式

In the first line output the only number — the minimum time the girl needs to put the objects into her handbag.

In the second line output the possible optimum way for Lena. Each object in the input is described by its index number (from 1 to nn ), the handbag‘s point is described by number 0. The path should start and end in the handbag‘s point. If there are several optimal paths, print any of them.

这道题暴力状压很好打，但是2^n*n^2的复杂度根本过不了。

但是你想想，对于所有操作的顺序是可以变的，答案不会变，因为我的状态中有这个东西，这东西不管是在什么时候捡的，它放到最后一次捡肯定没问题，于是对于任意在状态中的物品，我都可以用它来转移，所以我们任取一个物品，暴力枚举另外一个和它一起捡的，就可以通过此题了。

#include<bits/stdc++.h>
using namespace std;
int n,sx,sy,x[30],y[30],f[20000000],z[20000000];
int lowbit(int x){
    return x & -x;
} 
int dis(int u,int v){
    return (x[u]-x[v])*(x[u]-x[v])+(y[u]-y[v])*(y[u]-y[v]);
}
int H(int x){
    int cnt=0;
    while(x){
        cnt++;
        x>>=1;
    }
    return cnt;
} 
int main(){
    int i,j;
    cin>>x[0]>>y[0]>>n;
    for(i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);
    memset(f,0x3f,sizeof(f));
    f[0]=0;
    for(int state=1;state<(1<<n);state++){
        i=lowbit(state); 
        int w=state;
        while(w){
            j=lowbit(w);
            int nw=f[state^(i|j)]+dis(0,H(i))+dis(H(i),H(j))+dis(H(j),0);
            if(nw<f[state]){
                f[state]=nw;
                z[state]=state^(i|j);
            }
            w-=lowbit(w);
        }
    }
    cout<<f[(1<<n)-1]<<endl;
    int pd=(1<<n)-1;
    while(pd){
        printf("0 ");
        int t1=pd^z[pd];
        int t2=lowbit(pd);
        t1-=t2;
        printf("%d ",H(lowbit(t2)));
        if(t1)printf("%d ",H(lowbit(t1)));
        pd=z[pd];
    } 
    printf("0\n");
    return 0;
}