bzoj 3130 [Sdoi2013]费用流二分答案+网络流

最新推荐文章于 2019-02-06 10:33:32 发布

原创最新推荐文章于 2019-02-06 10:33:32 发布 · 203 阅读

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本文探讨了如何通过二分搜索求解最大流问题中流量最大的边的最小值，采用Dinic算法实现，并提供了详细的代码示例。

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题面

题目传送门

解法

题目转化一下就可以发现要求最大流中流过的流量最大的那条边流量最小
显然可以在实数域内二分答案 $m i d$ ，然后就变成所有边容量为 $m i n (c, m i d)$ 的情况下最大流是否不变
直接 $d i n i c$ 似乎就可以了，精度要求不高

代码

#include <bits/stdc++.h>
#define double long double
#define eps 1e-6
#define N 110
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
	int x, y, c;
} a[N * 10];
struct Edge {
	int next, num; double c;
} e[N * N * 5];
int n, m, p, cnt, l[N], cur[N];
void add(int x, int y, double c) {
	e[++cnt] = (Edge) {e[x].next, y, c};
	e[x].next = cnt;
}
void Add(int x, int y, double c) {
	add(x, y, c), add(y, x, 0);
}
bool bfs() {
	for (int i = 2; i <= n; i++) l[i] = -1;
	queue <int> q; q.push(1);
	while (!q.empty()) {
		int x = q.front(); q.pop();
		for (int p = e[x].next; p; p = e[p].next) {
			int k = e[p].num; double c = e[p].c;
			if (c > eps && l[k] == -1)
				l[k] = l[x] + 1, q.push(k);
		}
	}
	return l[n] != -1;
}
double dfs(int x, double lim) {
	if (x == n) return lim;
	double used = 0;
	for (int p = cur[x]; p; p = e[p].next) {
		int k = e[p].num; double c = e[p].c;
		if (c > eps && l[k] == l[x] + 1) {
			double w = dfs(k, min(lim - used, c));
			e[p].c -= w, e[p ^ 1].c += w, used += w;
			if (e[p].c > eps) cur[x] = p;
			if (lim - used < eps) return lim;
		}
	}
	if (used < eps) l[x] = -1; return used;
}
double dinic() {
	double ret = 0;
	while (bfs()) {
		for (int i = 1; i <= n; i++) cur[i] = e[i].next;
		ret += dfs(1, INT_MAX);
	}
	return ret;
}
bool check(double mid, double tmp) {
	cnt = n + (n % 2 == 0);
	for (int i = 1; i <= n; i++) e[i].next = 0;
	for (int i = 1; i <= m; i++) {
		int x = a[i].x, y = a[i].y; double c = a[i].c;
		if (mid > c) Add(x, y, c); else Add(x, y, mid);
	}
	return tmp - dinic() < eps;
}
int main() {
	read(n), read(m), read(p); cnt = n + (n % 2 == 0);
	for (int i = 1; i <= m; i++) {
		read(a[i].x), read(a[i].y), read(a[i].c);
		Add(a[i].x, a[i].y, a[i].c);
	}
	double tmp = dinic(); cout << tmp << "\n";
	double l = 0, r = tmp;
	while (l + eps < r) {
		double mid = (l + r) / 2;
		if (check(mid, tmp)) r = mid;
			else l = mid;
	}
	cout << fixed << setprecision(6) << r * p << "\n";
	return 0;
}