HDU 5313

要构成最大完全图，就意味着节点二分为两半乘积最大，那么该题目转化为，先对所有联通分量二分匹配，求出每个分量的两个二分值。然后求这些从每个两个里面选一个最终能够构成1-n中的哪些值，然后暴力一下即可。对于从m个二元组中每个选一个数都成最终数字，可以用普通的O（m*n）的背包但本题目n<=10000，用bitset优化背包，即可。

本体自己代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include <cmath>
#include <bitset>
#define clr(a, x) memset(a, x, sizeof a)
#define ALL(a) a.begin(), a.end()
#define ls (rt<<1)
#define rs (ls|1)
#define lson l, mid, ls
#define rson mid+1, r, rs
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;

const int maxn=10000+100;

int n, scc_cnt, col[maxn];
int scc[maxn][3];
bool d[maxn];
vector<int> G[maxn], ans[3];

void init()
{
    scc_cnt=0;
    memset(col, 0, sizeof col);
    memset(d, 0, sizeof d);
    ans[0].clear();
    ans[1].clear();
    clr(scc, 0);
    for(int i=0;i<=n;i++) G[i].clear();
}

bool dfs(int u)
{
    for(int i=0; i<G[u].size(); i++)
    {
        int v=G[u][i];
        if(col[u]==col[v])return false;
        if(!col[v])
        {
            col[v]=3-col[u];
            scc[scc_cnt][col[v]-1]++;
            if(!dfs(v))return false;
        }
    }
    return true;
}
 int T, m;
int dp()
{
  typedef bitset<10001> Bitset;
        Bitset cur;
        cur.set(0);
        for(int i=1;i<=scc_cnt;i++) {
            int a = scc[i][0], b = scc[i][1];
            if(a == 0 && b == 0) continue;
            cur = cur << a | cur << b;
        }
        long long ans = 0;
        for(int i=1;i<=n;i++) if(cur[i]){
            ans = max((long long )ans, (long long)i * (n-i) - m);
        }
        cout << ans << endl;
        return 0;
}

int main()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        init();
        for(int i=0; i<m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        for(int i=1; i<=n; i++)
            if(!col[i])
            {
                col[i]=1;
                scc[++scc_cnt][0]++;
                dfs(i);
            }
        dp();
    }
    return 0;
}

附anta代码：

#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#include <bitset>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

struct UnionFind {
    vector<int> data;
    void init(int n) { data.assign(n, -1); }
    bool unionSet(int x, int y) {
        x = root(x); y = root(y);
        if(x != y) {
            if(data[y] < data[x]) swap(x, y);
            data[x] += data[y]; data[y] = x;
        }
        return x != y;
    }
    bool findSet(int x, int y) { return root(x) == root(y); }
    int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
    int size(int x) { return -data[root(x)]; }
};

int main() {
    int T;
    scanf("%d", &T);
    rep(ii, T) {
        int n, m;
        scanf("%d%d", &n, &m);
        UnionFind uf, uf2;
        uf.init(n);
        uf2.init(n * 2);
        rep(i, m) {
            int u, v;
            scanf("%d%d", &u, &v), -- u, -- v;
            uf.unionSet(u, v);
            uf2.unionSet(u, n + v);
            uf2.unionSet(n + u, v);
        }
        vi cnt1(n, 0), cnt2(n, 0);
        rep(i, n) {
            if(uf2.findSet(i, uf.root(i)))
                ++ cnt1[uf.root(i)];
            else
                ++ cnt2[uf.root(i)];
        }
        typedef bitset<10001> Bitset;
        Bitset cur;
        cur.set(0);
        rep(i, n) {
            int a = cnt1[i], b = cnt2[i];
            if(a == 0 && b == 0) continue;
            cur = cur << a | cur << b;
        }
        ll ans = 0;
        rer(i, 0, n) if(cur[i])
            amax(ans, (ll)i * (n-i) - m);
        cout << ans << endl;
    }
    return 0;
}