HDU 3335（舞蹈鏈）

(http://acm.hust.edu.cn/vjudge/contest/view.action?cid=106836#problem/I)
題意：在n個數中，選出最多的數，使他們倆倆不能被整除。
解法：將問題轉化成刪除最少的數，使其滿足題意。若i可以被j整除，則i，j不能同時被選，使列a向i，j連邊。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <complex>
//#include <tr1/unordered_set>
//#include <tr1/unordered_map>
#include <bitset>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define inf (1e9+7)
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define clr(x, y) memset(x, y, sizeof x)
#define LL long long
#define ll long long
#define uLL unsigned LL

using namespace std;

const int maxn = 4e6+30;

struct sad{
    int L[maxn],R[maxn],U[maxn],D[maxn];
    int Row[maxn],Col[maxn];
    int h[maxn],s[maxn];
    int si,ans;
    void init(int m){
        for(int i=0;i<=m;i++){
            U[i] = D[i] = i;
            L[i] = i-1;
            R[i] = i+1;
            s[i] = 0;
        }
        L[0] = m;
        R[m] = 0;
        clr(h,-1);
        si=m;
        ans=inf;
    }
    void link(int row,int col){
        si++;
        Col[si] = col;
        Row[si] = row;
        s[col]++;
        U[si] = U[col];
        D[si] = col;
        D[U[col]] = si;
        U[col] = si;
        if( h[row] != -1 ){
            R[si] = h[row];
            L[si] = L[h[row]];
            R[L[si]] = si;
            L[R[si]] = si;
        }else
            h[row] = L[si] = R[si] = si;
    }
    void Remove(int c){
        for(int i=D[c];i-c;i=D[i]){
            L[R[i]] = L[i];
            R[L[i]] = R[i];
        }
    }
    void Resume(int c){
        for(int i=U[c];i-c;i=U[i]){
            L[R[i]] = R[L[i]] = i;
        }
    }
    int vis[maxn];
    int H(){
        int ret=0;
        for(int i=R[0];i;i=R[i])
            vis[i] = 0;
        for(int i=R[0];i;i=R[i]) if(!vis[i]){
            ret++;
            vis[i] = 1;
            for(int j=D[i];j-i;j=D[j])
                for(int k=R[j];k-j;k=R[k])
                    vis[Col[k]] = 1;
        }
        return ret;
    }
    void dfs(int cnt){
        if( cnt + H() >= ans ) return ;
        if( R[0] == 0 ){
            ans = min(ans,cnt);
            return ;
        }
        int c=R[0];
        for(int i=R[0];i;i=R[i])
            if(s[c]>s[i]) c=i;
        for(int i=D[c];i-c;i=D[i]){
            Remove(i);
            for(int j=R[i];j-i;j=R[j])
                Remove(j);
            dfs(cnt+1);
            for(int j=L[i];j!=i;j=L[j])
                Resume(j);
            Resume(i);
        }
    }
}dlx;

vector<pair<int,int> >Q;
ll a[maxn];

int main(){
//    freopen("input.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%I64d",&a[i]);
        int cnt = 0;
        Q.clear();
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++) if(a[i]%a[j]==0||a[j]%a[i]==0){
                ++cnt;
                Q.push_back( make_pair(i+1,cnt) );
                Q.push_back( make_pair(j+1,cnt) );
            }
        dlx.init(cnt);
        for(int i=0;i<Q.size();i++)
            dlx.link( Q[i].first , Q[i].second );
        dlx.dfs(0);
        printf("%d\n",n-dlx.ans);
    }
    return 0;
}