CodeForces 653D. Delivery Bears

Niwel is a little golden bear. As everyone knows, bears live in forests, but Niwel got tired of seeing all the trees so he decided to move to the city.

In the city, Niwel took on a job managing bears to deliver goods. The city that he lives in can be represented as a directed graph with nnodes and m edges. Each edge has a weight capacity. A delivery consists of a bear carrying weights with their bear hands on a simple path from node 1 to node n. The total weight that travels across a particular edge must not exceed the weight capacity of that edge.

Niwel has exactly x bears. In the interest of fairness, no bear can rest, and the weight that each bear carries must be exactly the same. However, each bear may take different paths if they like.

Niwel would like to determine, what is the maximum amount of weight he can deliver (it’s the sum of weights carried by bears). Find the maximum weight.

Input

The first line contains three integers n, m and x (2 ≤ n ≤ 50, 1 ≤ m ≤ 500, 1 ≤ x ≤ 100 000) — the number of nodes, the number of directed edges and the number of bears, respectively.

Each of the following m lines contains three integers ai, bi and ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 1 000 000). This represents a directed edge from node ai to bi with weight capacity ci. There are no self loops and no multiple edges from one city to the other city. More formally, for each i and j that i ≠ j it’s guaranteed that ai ≠ aj or bi ≠ bj. It is also guaranteed that there is at least one path from node 1 to node n.

Output

Print one real value on a single line — the maximum amount of weight Niwel can deliver if he uses exactly x bears. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let’s assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct if 《CodeForces 653D. Delivery Bears》.

Examples
input
4 4 3
1 2 2
2 4 1
1 3 1
3 4 2
output
1.5000000000
input
5 11 23
1 2 3
2 3 4
3 4 5
4 5 6
1 3 4
2 4 5
3 5 6
1 4 2
2 5 3
1 5 2
3 2 30
output
10.2222222222
Note

In the first sample, Niwel has three bears. Two bears can choose the path 《CodeForces 653D. Delivery Bears》, while one bear can choose the path 《CodeForces 653D. Delivery Bears》. Even though the bear that goes on the path 《CodeForces 653D. Delivery Bears》 can carry one unit of weight, in the interest of fairness, he is restricted to carry 0.5 units of weight. Thus, the total weight is 1.5 units overall. Note that even though Niwel can deliver more weight with just 2 bears, he must use exactly 3 bears on this day.

 

Solution

当时交的时候精度跪了,因为我用的是longlong的二分…今天才知道浮点数二分可以控制二分次数…
二分答案ans,检查的时候每条边的容量除以ans为该条边的流量,跑一边dinic,看看maxflow和熊的数量大小关系即可。
还要注意,c[i]/ans时,可能会溢出int,WA on test 45就是这个情况。
《CodeForces 653D. Delivery Bears》

Code

#include <cstdio>
#include <queue>
using namespace std;
int getint() {
	int r = 0; char c = getchar();
	for (; c < '0' || c > '9'; c = getchar());
	for (; '0' <= c && c <= '9'; c = getchar()) r = r * 10 - '0' + c;
	return r;
}
const int maxn = 510;
const int INF = 1000000007;
int S, T, cnte = 1;
int n, m;
long long sbj;
typedef long long ll;
int ea[maxn*10], eb[maxn*10], h[maxn*10];
double ec[maxn*10];
struct edge_type {
	int to, next;
	ll r;
} edge[1000005];
void ins(int u, int v, ll w) {
	edge[++cnte].to = v;
	edge[cnte].next = h[u];
	h[u] = cnte;
	edge[cnte].r = w;
	edge[++cnte].to = u;
	edge[cnte].next = h[v];
	h[v] = cnte;
	edge[cnte].r = 0;
}
void build(double x) {
	S = 1;
	T = n;
	for (int i = 1; i <= n; ++i) h[i] = 0;
	cnte = 1;
	for (int i = 1; i <= m; ++i) ins(ea[i], eb[i], ll(ec[i]/x));
}
int q[20100], head, tail;
int dis[20100];
bool bfs() {
	head = tail = 0;
	q[tail++] = S;
	for (int i = S; i <= T; ++i) dis[i] = -1;
	dis[S] = 0;
	int now;
	while(head < tail) {
		now = q[head++];
		for (int i = h[now]; i; i = edge[i].next) {
			if (dis[edge[i].to] == -1 && edge[i].r > 0) {
				dis[edge[i].to] = dis[now] + 1;
				q[tail++] = edge[i].to;
			}
		}
	}
	if (dis[T] == -1) return false;
	return true;
}
ll dfs(int now, ll flow) {
	if (flow == 0 || now == T) return flow;
	ll f, ret = 0;
	for (int i = h[now]; i; i = edge[i].next) {
		if (edge[i].r && dis[edge[i].to] == dis[now]+1)
			if(f=dfs(edge[i].to,min(flow,edge[i].r))) {
				ret += f;
				edge[i].r -= f;
				edge[i^1].r += f;
				flow -= f;
				if (flow == 0) break;
			}
	}
	if (ret == 0) dis[now] = -1;
	return ret;
}
ll dinic() {
	ll ret = 0;
	while (bfs())
		ret += dfs(S,INF);
	return ret;
}
bool check(double x) {
	build(x);
	int ans = dinic();
	if (ans >= sbj) return true;
	return false;
}
int main () {
	n = getint(); m = getint(); sbj = getint();
	for (int i = 1; i <= m; ++i) {
		ea[i] = getint();eb[i] = getint();ec[i] = getint();
		//ec[i] -= 0.0000005;
	}
	double l = 0.0000001, r = 1000000, mid;
	for (int i = 0; i < 70; ++i) {
		double mid = (l+r)/2;
		if (check(mid)) l = mid;
		else r = mid;
	}
	printf("%.9lf", l*sbj);
}

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