[01分数规划二分]网络战争

网络战争

给一个无向图 (G=(V,E)) ，求一个边集 (C) ，删除 (C) 后 (s) ，(t) 不再连通

最小化

[dfrac {sum_{ein C}w_e} {|C|} ]

01分数规划

每个物品价值 (w_i) , 花费 (c_i)

选一些物品使得

(dfrac{sum w} {sum c}) 最小

[dfrac {sum w} {sum c} < lambda \ sum w - lambda sum c < 0 \ sum_i (w_i - lambda c_i) < 0 ]

至此，可以二分答案 (lambda)

/*
 * @Author: zhl
 * @Date: 2020-10-22 15:44:57
 */

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i = a;i <= b;i++)
#define repE(i,u) for(int i = head[u]; ~i; i = E[i].next)
using namespace std;

const int N = 2100, M = 10100, inf = 1e9;
struct Edge {
	int to, next;
	double flow;
}E[M << 1];
int head[N], tot;
void addEdge(int from, int to, double w) {
	E[tot] = Edge{ to,head[from],w };
	head[from] = tot++;
	E[tot] = Edge{ from,head[to],0.0 };
	head[to] = tot++;
}

int n, m, s, t;
queue<int>Q;
int dis[N], cur[N];
bool bfs() {
	memset(dis, -1, sizeof dis);
	while (!Q.empty())Q.pop();
	Q.push(s); dis[s] = 0; cur[s] = head[s];
	while (!Q.empty()) {
		int u = Q.front(); Q.pop();
		repE(i, u) {
			int v = E[i].to;
			if (dis[v] == -1 and E[i].flow) {
				cur[v] = head[v];
				dis[v] = dis[u] + 1;
				Q.push(v);
				if (v == t)return true;
			}
		}
	}
	return false;
}

double dfs(int u, double limit) {
	if (u == t)return limit;
	double k, res = 0;
	for (int i = cur[u]; ~i and res < limit; i = E[i].next) {
		int v = E[i].to;
		cur[u] = i;
		if (dis[v] == dis[u] + 1 and E[i].flow) {
			k = dfs(v, min(limit - res, E[i].flow));
			if (k == 0)dis[v] = -1;
			E[i].flow -= k; E[i ^ 1].flow += k;
			res += k;
		}
	}
	return res;
}

double Dinic() {
    double res = 0, f;
	while (bfs())while (f = dfs(s, inf)) res += f;
	return res;
}

const double eps = 1e-4;
int u[N],v[N],w[N];

bool judge(double x){
    memset(head,-1,sizeof head);
    tot = 0;
    double sum = 0;
    rep(i,1,m){
        if(w[i] <= x) sum += w[i] - x;
        else addEdge(u[i],v[i],w[i]-x),addEdge(v[i],u[i],w[i]-x);
    }
    return Dinic() + sum <= 0;
}
int main(){
    scanf("%d%d%d%d",&n,&m,&s,&t);
    rep(i,1,m){
        scanf("%d%d%d",u+i,v+i,w+i);
    }
    double l = 0,r = 1e7;
    while(r - l > eps){
        double mid = (l + r) / 2;
        if(judge(mid))r = mid;
        else l = mid;
    }
    printf("%.2f
",l);
}