codeforces 8C. Looking for Order 状压dp

题目链接

给n个物品的坐标， 和一个包裹的位置， 包裹不能移动。 每次最多可以拿两个物品， 然后将它们放到包里， 求将所有物品放到包里所需走的最小路程。

直接状压dp就好了。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int n;
int dp[1<<24], dis[25][25], b[1<<24], ans[500], cnt;
pll a[25];
int main()
{
    int x, y;
    cin>>a[0].fi>>a[0].se;
    cin>>n;
    a[n] = a[0];
    for(int i = 0; i<n; i++) {
        scanf("%d%d", &a[i].fi, &a[i].se);
    }
    for(int i = 0; i<=n; i++) {
        for(int j = 0; j<=n; j++) {
            dis[i][j] = (a[i].fi-a[j].fi)*(a[i].fi-a[j].fi)+(a[i].se-a[j].se)*(a[i].se-a[j].se);
        }
    }
    mem2(dp);
    dp[0] = 0;
    for(int i = 0; i<(1<<n); i++) {
        if(dp[i]==inf)
            continue;
        for(int j = 0; j<n; j++) {
            if((1<<j&i)==0) {
                for(int k = j; k<n; k++) {
                    if((1<<k&i)==0) {
                        int tmp = i|(1<<j)|(1<<k);
                        int d = dis[j][k]+dis[n][j]+dis[n][k];
                        if(dp[tmp]>dp[i]+d) {
                            dp[tmp] = dp[i]+d;
                            b[tmp] = i;
                        }
                    }
                }
                break;
            }
        }
    }
    cout<<dp[(1<<n)-1]<<endl;
    for(int i = (1<<n)-1; i!=0; i = b[i]) {
        int tmp = b[i]^i;
        ans[cnt++] = 0;
        for(int j = 0; j<n; j++) {
            if((1<<j)&tmp) {
                ans[cnt++] = j+1;
            }
        }
    }
    ans[cnt++] = 0;
    for(int i = cnt-1; i>=0; i--) {
        printf("%d ", ans[i]);
    }
    return 0;
}