Bzoj3597: [Scoi2014]方伯伯运椰子

题面

传送门

Sol

消圈定理：如果一个费用流网络的残量网络有负环，那么这个费用流不优
于是这个题就可以建出残量网络，然后分数规划跑负环了

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int maxn(6005);
const double eps(1e-5);

int n, m, first[maxn], cnt, vis[maxn];
double dis[maxn], l = 0, r = 5e4;

struct Edge{
	int to, next;
	double w;
} edge[maxn];

IL void Add(RG int u, RG int v, RG double w){
	edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}

IL int Dfs(RG int u, RG double w){
	vis[u] = 1;
	for(RG int e = first[u]; e != -1; e = edge[e].next){
		RG int v = edge[e].to;
		RG double d = dis[u] + edge[e].w + w;
		if(dis[v] > d){
			dis[v] = d;
			if(vis[v] || Dfs(v, w)) return 1;
		}
	}
	vis[u] = 0;
	return 0;
}

IL int Check(RG double v){
	for(RG int i = 1; i <= n + 2; ++i) dis[i] = 0, vis[i] = 0;
	for(RG int i = 1; i <= n + 2; ++i) if(Dfs(i, v)) return 1;
	return 0;
}

int main(){
	n = Input(), m = Input();
	for(RG int i = 1; i <= n + 2; ++i) first[i] = -1;
	for(RG int i = 1; i <= m; ++i){
		RG int u = Input(), v = Input(), a = Input(), b = Input(), c = Input(), d = Input();
		Add(u, v, b + d);
		if(c) Add(v, u, a - d);
	}
	while(r - l >= eps){
		RG double mid = (l + r) / 2.0;
		if(Check(mid)) l = mid;
		else r = mid;
	}
	printf("%.2lf\n", r);
	return 0;
}