codeforces 8C 状压DP

codeforces 8C

---

题意：

$给定n个物品，第i个物品的坐标为(x_i,y_i)，你的起点坐标为(x_s,y_s)。$
$现在要将所有物品全部放回起点，每次手中拿的物品数量不能超过2个。$
$从一个位置到另外一个位置花费的时间为距离的平方。问最少时间。$

---

题解：

$dp[i[代表i状态花费的最少时间。$

• $dp[i|1<<j|1<<k]=min(dp[i|1<<j|1<<k],dp[i]+dis[0][j+1]+dis[j+1][k+1]+dis[k+1][0])$

---

#include 
using namespace std;
int x[25], y[25], dis[25][25], path[1<<24];
int dp[1<<24];

int main() {
    int n;
    cin >> x[0] >> y[0] >> n;
    for(int i = 1 ; i <= n ; i++){
        cin >> x[i] >> y[i];
    }
    for(int i = 0 ; i <= n ; i++){
        for(int j = 0 ; j <= n ; j++){
            dis[i][j] = (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
        }
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0] = 0;
    for(int i = 0 ; i < (1<<n)-1 ; i++){
        if(dp[i] < 0x3f3f3f3f){
            for(int j = 0 ; j < n ; j++){
                if(!(i>>j & 1)){
                    for(int k = j ; k < n ; k++){
                        if(!(i>>k & 1)){
                            if(dp[i | 1<<j | 1<<k] > dp[i]+dis[0][j+1]+dis[j+1][k+1]+dis[k+1][0]){
                                dp[i | 1<<j | 1<<k] = dp[i]+dis[0][j+1]+dis[j+1][k+1]+dis[k+1][0];
                                path[i | 1<<j | 1<<k] = i;
                            }  
                        }
                    }
                    break;
                }
            }
        }
    }
    cout << dp[(1<<n)-1] << endl;;
    cout << 0;
    for(int i = (1<<n)-1 ; i > 0 ; i = path[i]){
        for(int j = i^path[i] ; j > 0 ; j -= j & -j){
            cout << ' ' << (int)(log(j&-j)/log(2)+1e-8+1);
        }
        cout << ' ' << 0;
    }
    return 0; 
}