bzoj 3597: [Scoi2014]方伯伯运椰子 [01分数规划 消圈定理 spfa负环]

3597: [Scoi2014]方伯伯运椰子

题意：

from mhy12345

给你一个满流网络，对于每一条边，压缩容量1 需要费用ai，扩展容量1 需要bi，
当前容量上限ci，每单位通过该边花费di，限制网络流量不能改变。调整后必须满
流，设调整了K 次，使得费用减少量为D，最大化D/K

---

就是给你一个费用流，但不是最小，增广的费用为b+d，退流的费用为a-d

就是正反向增广路

根据消圈定理，流f为mcmf当且仅当无负费用增广圈

01分数规划+spfa求负环即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N = 505, M = 1e4+5, mo = 1e9+7;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m;
double l, r;

struct edge {int v, ne; double w;} e[M];
int cnt, h[N];
inline void ins(int u, int v, double w) {
	e[++cnt] = (edge) {v, h[u], w}; h[u] = cnt;
}
namespace nc {
	int vis[N], T; double d[N];
	bool dfs(int u, double mid) {
		vis[u] = T;
		for(int i=h[u]; i; i=e[i].ne) {
			int v = e[i].v; double w = e[i].w + mid;
			if(d[v] > d[u] + w) {
				d[v] = d[u] + w;
				if(vis[v] == T || dfs(v, mid)) return true;
			}
		}
		vis[u] = 0;
		return false;
	}
	bool neg_circle(double mid) {
		memset(vis, 0, sizeof(vis));
		memset(d, 0, sizeof(d));
		for(T=1; T<=n+2; T++) if(dfs(T, mid)) return true;
		return false;
	}
}
bool check(double mid) {
	return nc::neg_circle(mid);
}
int main() {
	freopen("in", "r", stdin);
	n = read(); m = read();
	for(int i=1; i<=m; i++) {
		int u = read(), v = read(), a = read(), b = read(), c = read(), d = read();
		ins(u, v, b+d); if(c) ins(v, u, a-d);
		r += c * d;
	}
	while(r - l > 1e-3) {
		double mid = (l+r)/2.0;
		if(check(mid)) l = mid;
		else r = mid;
	}
	printf("%.2lf
", l);
}