[CF] 8C Looking for Order

状压模板题 CF难度2000？ 我得好好了解一下CF的难度机制了

反正CF的难度比洛谷真实就好了

Code

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define N 25
#define INF 0x3f3f3f3f
using namespace std;
inline int read() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
    return x * f;
}
int n;
int x[N],y[N];
int g[N][N];
int f[1<<N],pre[1<<N];  //状压dp 
void print(int u) {
    printf("0 ");
    if(u == 0) return;
    int bit = u ^ pre[u];   //拿出这两个数 
    for(int i=1;i<=n;++i)
        if(bit & (1<<(i-1)))
            printf("%d ",i);
    print(pre[u]);
}
int main()
{
    x[0] = read(), y[0] = read(); n = read();
    for(int i=1;i<=n;++i)
        x[i] = read(), y[i] = read();
    memset(g, 0x3f, sizeof(g));
    for(int i=0;i<=n;++i)
        for(int j=0;j<=n;++j)
            g[i][j] = g[j][i] = (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]);
    memset(f, 0x3f, sizeof(f));
    int maxn = (1<<n)-1;
    f[0] = 0;
    for(int S=0;S<=maxn;++S) {  //state
        if(f[S] == INF) continue;
        for(int i=1;i<=n;++i) {
            if(S & (1<<(i-1))) continue;
            for(int j=1;j<=n;++j) {
                if(S & (1<<(j-1))) continue;
                int x = S | (1<<(i-1)) | (1<<(j-1));
                int y = f[S] + g[0][i] + g[i][j] + g[j][0];
                if(y < f[x]) {
                    f[x] = y;
                    pre[x] = S;
                }
            }
            break;
        }
    }
    printf("%d\n",f[maxn]);
    print(maxn);
    return 0;
}

[CF] 8C Looking for Order

原文：https://www.cnblogs.com/BaseAI/p/11741229.html