题目链接

BZOJ3597

题解

orz一眼过去一点思路都没有

既然是流量网络，就要借鉴网络流的思想了

我们先处理一下那个比值，显然是一个分数规划，我们二分一个\(\lambda = \frac{X - Y}{k}\)
如果\(\lambda\)成立，则

\[\lambda \le \frac{X - Y}{k} \]

即

\[\lambda k + (Y - X) \le 0 \]

所以我们只需要判断是否存在一种方案使得这个式子成立

依照网络流的思想，撤回流量就往反向边走，扩展流量往正向边
对于边\((u,v)\)，撤回流量产生的代价就是\(\lambda + \Delta fee = \lambda + (ai - di)\)
扩展产生的代价就是\(\lambda + \Delta fee = \lambda + (bi + di)\)
为保证流量守恒，我们调整边走过的路径必须成环
如果我们建出的图中存在权值为负的环，就找到了一条满足上式的调整方案
使用\(spfa\)判负环即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define eps 1e-4
using namespace std;
const int maxn = 5005,maxm = 10005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne;
struct EDGE{int to,nxt; double w;}ed[maxm];
inline void build(int u,int v,double w){
	ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
}
int n,m,vis[maxn];
double d[maxn];
bool spfa(int u,double lam){
	vis[u] = true;
	Redge(u){
		to = ed[k].to;
		if (d[to] > d[u] + ed[k].w + lam){
			d[to] = d[u] + ed[k].w + lam;
			if (vis[to]) return true;
			if (spfa(to,lam)) return true;
		}
	}
	vis[u] = false;
	return false;
}
bool check(double lam){
	cls(vis); cls(d);
	for (int i = 1; i <= n; i++) if (spfa(i,lam)) return true;
	return false;
}
int main(){
	n = read() + 2; m = read();
	int u,v,a,b,c,d;
	for (int i = 1; i <= m; i++){
		u = read(); v = read(); a = read(); b = read(); c = read(); d = read();
		if (u == n - 1 || v == n - 1) continue;
		if (c) build(v,u,a - d);
		build(u,v,d + b);
	}
	double l = 0,r = 1000000000,mid;
	while (r - l > eps){
		mid = (l + r) / 2.0;
		if (check(mid)) l = mid;
		else r = mid;
	}
	printf("%.2lf",(l + r) / 2.0);
	return 0;
}