| Time Limit: 8000MS | Memory Limit: 65536K | |
| Total Submissions: 6908 | Accepted: 1998 | |
| Case Time Limit: 2000MS | Special Judge |
Description
John is a Chief Executive Officer at a privately owned medium size company. The owner of the company has decided to make his son Scott a manager in the company. John fears that the owner will ultimately give CEO position to Scott if he does well on his new manager position, so he decided to make Scott’s life as hard as possible by carefully selecting the team he is going to manage in the company.
John knows which pairs of his people work poorly in the same team. John introduced a hardness factor of a team — it is a number of pairs of people from this team who work poorly in the same team divided by the total number of people in the team. The larger is the hardness factor, the harder is this team to manage. John wants to find a group of people in the company that are hardest to manage and make it Scott’s team. Please, help him.
In the example on the picture the hardest team consists of people 1, 2, 4, and 5. Among 4 of them 5 pairs work poorly in the same team, thus hardness factor is equal to 5⁄4. If we add person number 3 to the team then hardness factor decreases to 6⁄5.
Input
The first line of the input file contains two integer numbers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 1000). Here n is a total number of people in the company (people are numbered from 1 to n), and m is the number of pairs of people who work poorly in the same team. Next m lines describe those pairs with two integer numbers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) on a line. The order of people in a pair is arbitrary and no pair is listed twice.
Output
Write to the output file an integer number k (1 ≤ k ≤ n) — the number of people in the hardest team, followed by k lines listing people from this team in ascending order. If there are multiple teams with the same hardness factor then write any one.
Sample Input
sample input #1 5 6 1 5 5 4 4 2 2 5 1 2 3 1 sample input #2 4 0
Sample Output
sample output #1 4 1 2 4 5 sample output #2 1 1
Hint
Note, that in the last example any team has hardness factor of zero, and any non-empty list of people is a valid answer.
題意：給出一個無向圖，n個點，m條邊，選一個子圖，使得該子圖 邊數：點數最大。
思路：最大密度子圖。
設原點s，匯點t。
原點s和1~n點連接，容量爲m。
對於原圖中的無向邊a，b，連接a->b，容量1，連接b->a，容量1。
二分枚舉一個double數字，左邊界L = 0，右邊界R = m。
跳出條件是while(R – L >= (1 / n / n))
對於每一個mid = (L + R) * 0.5。
1~n每個點和t連接，容量爲m + 2 * mid – d[i]。
求最大流得到一個double數字ans。
如果 (m * n – ans) * 0.5大於0， L = mid
否則R = mid。
最後跳出的L就是最大密度。
拿這個L再重新建圖，求最大流。
然後從s出發dfs，走殘留容量大於0的邊，所有能到達的點就是答案。
AC代碼：
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#include <ctime>
#include <vector>
#include <algorithm>
#define ll long long
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
using namespace std;
const double INF = 1e9;
const int maxn = 105;
const double eps = 1e-7;
struct Edge{
int u, v;
double cap, flow;
int next;
}et[maxn * maxn];
int cnt[maxn], dis[maxn], pre[maxn], cur[maxn], eh[maxn], ans[maxn];
double low[maxn];
int n, m, s, t, num, tot;
int a[1005], b[1005], d[maxn];
bool vis[maxn];
void init(){
memset(eh, -1, sizeof(eh));
num = 0;
}
void add(int u, int v, double cap, double flow){
Edge e = {u, v, cap, flow, eh[u]};
et[num] = e;
eh[u] = num++;
}
void addedge(int u, int v, double cap){
add(u, v, cap, 0);
add(v, u, 0, 0);
}
double isap(int s, int t, int nv){
int u, v, now;
double flow = 0;
memset(low, 0, sizeof(low));
memset(cnt, 0, sizeof(cnt));
memset(dis, 0, sizeof(dis));
for(u = 0; u <= nv; u++) cur[u] = eh[u];
low[s] = INF, cnt[0] = nv, u = s;
while(dis[s] < nv)
{
for(now = cur[u]; now != -1; now = et[now].next)
if(et[now].cap - et[now].flow > eps && dis[u] == dis[v = et[now].v] + 1) break;
if(now != -1)
{
cur[u] = pre[v] = now;
low[v] = min(low[u], et[now].cap - et[now].flow);
u = v;
if(u == t)
{
for(; u != s; u = et[pre[u]].u)
{
et[pre[u]].flow += low[t];
et[pre[u]^1].flow -= low[t];
}
flow += low[t];
low[s] = INF;
}
}
else
{
if(--cnt[dis[u]] == 0) break;
dis[u] = nv, cur[u] = eh[u];
for(now = eh[u]; now != -1; now = et[now].next)
if(et[now].cap - et[now].flow > eps&& dis[u] > dis[et[now].v] + 1)
dis[u] = dis[et[now].v] + 1;
cnt[dis[u]]++;
if(u != s) u = et[pre[u]].u;
}
}
return flow;
}
void build(double cc){
init();
for(int i = 1; i <= n; i++)
{
addedge(s, i, m);
addedge(i, t, m + 2 * cc - d[i]);
}
for(int i = 0; i < m; i++)
{
addedge(a[i], b[i], 1);
addedge(b[i], a[i], 1);
}
}
void dfs(int u){
vis[u] = 1;
if(u >= 1 && u <= n) ans[tot++] = u;
for(int i = eh[u]; i != -1; i = et[i].next)
if(et[i].cap - et[i].flow > eps)
{
int v = et[i].v;
if(!vis[v]) dfs(v);
}
}
int main()
{
while(~scanf("%d%d", &n, &m))
{
memset(d, 0, sizeof(d));
s = 0, t = n + 1;
for(int i = 0; i < m; i++)
{
scanf("%d%d", &a[i], &b[i]);
d[a[i]]++;
d[b[i]]++;
}
double low = 0, high = m, mmin = 1.0 / n / n;
while(high - low >= mmin)
{
double mid = (low + high) * 0.5;
build(mid);
if((m * n - isap(s, t, t + 1)) * 0.5 > 0) low = mid;
else high = mid;
}
build(low);
isap(s, t, t + 1);
tot = 0;
memset(vis, 0, sizeof(vis));
dfs(s);
if(tot == 0) ans[tot++] = 1;
sort(ans, ans + tot);
printf("%d\n", tot);
for(int i = 0; i < tot; i++) printf("%d\n", ans[i]);
}
return 0;
}