【SCOI2014】方伯伯运椰子

题面

https://loj.ac/problem/2214

一道好题。

最小费用最大流+调整法（$orz$杨颙）（原谅我想到杨颙，好像我们好久没见了）

先假设所有的路都被压缩至$w=0$，然后建两类边，用最小费用最大流扩大（类似的思想：棋盘占领）。

注意求出的东西应该是后来的费用$Y - $之前的费用$X + k \times mid$，所以$check()$的时候要取反（这个地方我想了好久。。。。。）

#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
#include<cmath>
#define ri register int
#define N 5050
#define M 3050
#define INF 1000000007
#define LL long long
#define T (n+2)
#define S (n+1)

using namespace std;

int n,m;
int u[M],v[M],a[M],b[M],c[M],d[M];

struct graph {
  vector<int> to,w,ed[N];
  vector<double> c;
  double dis[N]; int cur[N]; bool vis[N];
  void clear() {
    c.clear(); to.clear(); w.clear();
    for (ri i=1;i<=n+2;i++) ed[i].clear();
  }
  void add_edge(int a,int b,int aw,double ac) {
    to.push_back(b); w.push_back(aw); c.push_back(ac);  ed[a].push_back(to.size()-1);
    to.push_back(a); w.push_back(0);  c.push_back(-ac); ed[b].push_back(to.size()-1);
  }
  bool spfa() {
    for (ri i=1;i<=n+2;i++) dis[i]=100000000000000.7;
    memset(vis,0,sizeof(vis));
    queue<int> q;
    dis[S]=0;q.push(S);vis[S]=1;
    while (!q.empty()) {
      int x=q.front(); q.pop();
      for (ri i=0;i<ed[x].size();i++) {
        int e=ed[x][i];
        if (dis[to[e]]>dis[x]+c[e] && w[e]) {
          dis[to[e]]=dis[x]+c[e];
          if (!vis[to[e]]) vis[to[e]]=1,q.push(to[e]);
        }
      }
      vis[x]=0;
    }
    return dis[T]<100000000000000.7;
  }
  int dfs(int x,int lim) {
    if (x==T || !lim) return lim;
    LL sum=0; vis[x]=1;
    for (ri &i=cur[x];i<ed[x].size();i++) {
      int e=ed[x][i];
      if (fabs(dis[x]+c[e]-dis[to[e]])<1e-5 && w[e] && !vis[to[e]]) {
        int f=dfs(to[e],min(lim,w[e]));
        w[e]-=f; w[1^e]+=f;
        lim-=f; sum+=f;
        if (!lim) return sum;
      }
    }
    return sum;
  }
  double zkw() {
    double ret=0;
    while (spfa()) {
      memset(vis,0,sizeof(vis));
      memset(cur,0,sizeof(cur));
      ret+=dfs(S,INF)*dis[T];
    }
    return ret;
  }
} G;

bool can(double mid) {
  double dec=0;
  for (ri i=1;i<=m;i++) dec+=(a[i]-d[i]+mid)*c[i];
  for (ri i=1;i<=m;i++) {
    G.add_edge(u[i],v[i],c[i],-(a[i]+mid-d[i]));
    if (u[i]!=S) G.add_edge(u[i],v[i],INF,b[i]+mid+d[i]);
  }
  if (0-(dec+G.zkw())>0) return 1; else return 0;
}

int main() {
  scanf("%d %d",&n,&m);
  for (ri i=1;i<=m;i++) {
    scanf("%d %d %d %d %d %d",&u[i],&v[i],&a[i],&b[i],&c[i],&d[i]);
  }
  double lb=0,rb=50000.0;
  double ans=0;
  while (rb-lb>1e-3) {
    double mid=(lb+rb)/2;
    //printf("%.5lf %.5lf %.5lf\n",lb,rb,mid);
    G.clear();
    if (can(mid)) lb=ans=mid; else rb=mid;
  }
  printf("%.2lf\n",ans);
  return 0;
}