POJ3155 Hard Life

嘟嘟嘟
洛谷题面


看到比值，就能想到01分数规划。
令\(x = \frac{\sum{m_i}}{\sum{n_i}}\)，变一下：\(\sum{m_i - n_i * x} = 0\)。但是建图我就是看题解的了。
我们把每一条边也看成一个点，从源点想这个点连一条边权为\(1\)的边，然后对于这条边连接着的节点\(u, v\)，分别向\(u, v\)连一条容量为\(INF\)的边。然后从原图的每一个点向汇点连一条容量为\(x\)的边。
需要注意的是二分的精度应该是\(\frac{1}{n ^ 2}\)，如果是自己设的\(eps\)的话在某谷上会\(WA\)。有一个大佬的博客讲了是怎么来的，然而我没怎么看懂……

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const db INF = 1e10;
const db eps = 1e-8;
const int maxn = 105;
const int maxm = 1e3 + 5;
const int maxe = 6505;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, s, t;
struct Node
{
  int x, y;
}a[maxm];

struct Edge
{
  int nxt, from, to; db cap, flow;
}e[maxe];
int head[maxn + maxm], ecnt = -1;
void addEdge(int x, int y, db w)
{
  e[++ecnt] = (Edge){head[x], x, y, w, 0};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], y, x, 0, 0};
  head[y] = ecnt;
}
int dis[maxn + maxm];
bool bfs()
{
  Mem(dis, 0); dis[s] = 1;
  queue<int> q; q.push(s);
  while(!q.empty())
    {
      int now = q.front(); q.pop();
      for(int i = head[now], v; i != -1; i = e[i].nxt)
    {
      v = e[i].to;
      if(!dis[v] && e[i].cap > e[i].flow + eps)
        {
          dis[v] = dis[now] + 1;
          q.push(v);
        }
    }
    }
  return dis[t];
}
int cur[maxn + maxm];
db dfs(int now, db res)
{
  if(now == t || res < eps) return res;
  db flow = 0, f;
  for(int& i = cur[now], v; i != -1; i = e[i].nxt)
    {
      v = e[i].to;
      if(dis[v] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > eps)
    {
      e[i].flow += f; e[i ^ 1].flow -= f;
      flow += f; res -= f;
      if(res < eps) break;
    }   
    }
  return flow;
}

db maxflow()
{
  db flow = 0;
  while(bfs())
    {
      memcpy(cur, head, sizeof(head));
      flow += dfs(s, INF);
    }
  return flow;
}

bool judge(db x)
{
  Mem(head, -1);
  ecnt = -1;
  for(int i = 1; i <= m; ++i)
    {
      addEdge(n + i, a[i].x, INF);
      addEdge(n + i, a[i].y, INF);
    }
  for(int i = 1; i <= n; ++i) addEdge(i, t, x);
  for(int i = 1; i <= m; ++i) addEdge(s, n + i, 1);
  return (db)m - maxflow() > eps;
}

int main()
{
  while(scanf("%d%d", &n, &m) != EOF)
    {
      s = 0; t = n + m + 1;
      for(int i = 1; i <= m; ++i) a[i].x = read(), a[i].y = read();
      if(!m) {printf("1\n1\n\n"); continue;}
      db L = 0, R = m;
      db Eps = 1.00/ (db)n / (db)n;
      while(R - L > Eps)
    {
      db mid = (L + R) / 2;
      if(judge(mid)) L = mid;
      else R = mid;
    }
      judge(L); bfs();
      int ans = 0;
      for(int i = 1; i <= n; ++i) if(dis[i]) ans++;
      write(ans), enter;
      for(int i = 1; i <= n; ++i) if(dis[i]) write(i), enter;
      enter;
    }
  return 0;
}