二分图点染色 BestCoder 1st Anniversary($) 1004 Bipartite Graph

题目传送门

/*
　　二分图点染色：这题就是将点分成两个集合就可以了，点染色用dfs做， 剩下的点放到点少的集合里去
    官方解答：首先二分图可以分成两类点X和Y, 完全二分图的边数就是|X|*|Y|.我们的目的是max{|X|*|Y|}, 并且|X|+|Y|=n.
    修正：实现多连通块染色，然后贪心选择，将两个集合个数差大的连通块优先添加，能尽量使得un*vn最大
 */
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
using namespace std;

const int MAXN = 1e4 + 10;
const int MAXM = 1e5 + 10;
const int INF = 0x3f3f3f3f;
vector<int> G[MAXN];
int col[2*MAXN];
bool vis[MAXN];
struct Block    {       //连通块
    int u, v;
    bool operator < (const Block &r) const  {
        return u - v > r.u - r.v;
    }
}b[MAXN];
int n, m, un, vn;

void DFS(int u, int c)    {
    col[c]++; vis[u] = true;
    for (int i=0; i<G[u].size (); ++i)  {
        int v = G[u][i];
        if (vis[v]) continue;
        DFS (v, c ^ 1);
    }
}

int main(void)  {       //BestCoder 1st Anniversary($) 1004 Bipartite Graph
    //freopen ("D.in", "r", stdin);

    int T;  scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d", &n, &m);
        for (int i=1; i<=n; ++i)    G[i].clear ();
        for (int i=1; i<=m; ++i)    {
            int u, v;   scanf ("%d%d", &u, &v);
            G[u].push_back (v); G[v].push_back (u);
        }

        int color = 0;
        memset (vis, false, sizeof (vis));
        memset (col, 0, sizeof (col));
        for (int i=1; i<=n; ++i)    {
            if (vis[i]) continue;
            DFS (i, color);    color += 2;
        }
        int cnt = 0;
        for (int i=0; i<color; i+=2) {
            b[++cnt].u = col[i];  b[cnt].v = col[i^1];
            if (b[cnt].u < b[cnt].v)    swap (b[cnt].u, b[cnt].v);
        }
        sort (b+1, b+1+cnt);

        un = vn = 0;
        for (int i=1; i<=cnt; ++i)  {
            if (un <= vn)   {
                un += b[i].u;   vn += b[i].v;
            }
            else    {
                un += b[i].v;   vn += b[i].u;
            }
        }
        printf ("%d
", un * vn - m);
    }

    return 0;
}