生活的艰辛（最小割，最大密度子图）

题意

求不带点权和边权的最大密度子图

思路

最大密度子图模板题

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

#define x first
#define y second

using namespace std;

typedef pair<int, int> pii;

const int N = 110, M = 2410;
const double eps = 1e-8, inf = 1e8;

int n, m, S, T;
int h[N], e[M], ne[M], idx;
double f[M];
int cur[N], d[N];
int dg[N];
bool st[N];
int ans;
pii edge[M];

void add(int a, int b, double c1, double c2)
{
    e[idx] = b, f[idx] = c1, ne[idx] = h[a], h[a] = idx ++;
    e[idx] = a, f[idx] = c2, ne[idx] = h[b], h[b] = idx ++;
}

void build(double mid)
{
    memset(h, -1, sizeof(h));
    idx = 0;
    for(int i = 0; i < m; i ++) {
        int a = edge[i].x, b = edge[i].y;
        add(a, b, 1, 1);
    }
    for(int i = 1; i <= n; i ++) {
        add(S, i, m, 0);
        add(i, T, 2 * mid - dg[i] + m, 0);
    }
}

bool bfs()
{
    memset(d, -1, sizeof(d));
    queue<int> que;
    que.push(S);
    d[S] = 0, cur[S] = h[S];
    while(que.size()) {
        int t = que.front();
        que.pop();
        for(int i = h[t]; ~i; i = ne[i]) {
            int ver = e[i];
            if(d[ver] == -1 && f[i] > 0) {
                d[ver] = d[t] + 1;
                cur[ver] = h[ver];
                if(ver == T) return true;
                que.push(ver);
            }
        }
    }
    return false;
}

double find(int u, double limit)
{
    if(u == T) return limit;
    double flow = 0;
    for(int i = cur[u]; ~i && limit > flow; i = ne[i]) {
        cur[u] = i;
        int ver = e[i];
        if(d[ver] == d[u] + 1 && f[i] > 0) {
            double t = find(ver, min(f[i], limit - flow));
            if(t <= 0) d[ver] = -1;
            f[i] -= t, f[i ^ 1] += t, flow += t;
        }
    }
    return flow;
}

double dinic(double mid)
{
    build(mid);
    double res = 0, flow;
    while(bfs()) {
        while(flow = find(S, inf)) {
            res += flow;
        }
    }
    return res;
}

void dfs(int u)
{
    st[u] = true;
    if(u != S) ans ++;
    for(int i = h[u]; ~i; i = ne[i]) {
        int ver = e[i];
        if(!st[ver] && f[i] > 0) {
            dfs(ver);
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    S = 0, T = n + 1;
    for(int i = 0; i < m; i ++) {
        int a, b;
        scanf("%d%d", &a, &b);
        edge[i] = {a, b};
        dg[a] ++, dg[b] ++;
    }
    double l = 0, r = 1000;
    while(r - l > eps) {
        double mid = (l + r) / 2;
        double res = dinic(mid);
        if(n * m - res > 0) l = mid;
        else r = mid;
    }
    dinic(l);
    dfs(S);
    if(!ans) puts("1
1");
    else {
        printf("%d
", ans);
        for(int i = 1; i <= n; i ++) {
            if(st[i]) {
                printf("%d
", i);
            }
        }
    }
    return 0;
}