codeforces 653D. Delivery Bears 网络流

题目链接

我们二分每个人携带的数量， 然后每个边的容量就相当于min(权值/二分的值, x).　ｘ是人的数量。
然后判断是否满流就可以。

这么裸的网络流为竟然没看出来。

注意写fsbs(r-l)>eps会挂掉...

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 1e5;
int q[maxn*2], head[maxn*2], dis[maxn/10], s, t, num;
struct node
{
    int to, nextt, c;
    node(){}
    node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
}e[maxn*2];
void init() {
    num = 0;
    mem1(head);
}
void add(int u, int v, int c) {
    e[num] = node(v, head[u], c); head[u] = num++;
    e[num] = node(u, head[v], 0); head[v] = num++;
}
int bfs() {
    mem(dis);
    dis[s] = 1;
    int st = 0, ed = 0;
    q[ed++] = s;
    while(st<ed) {
        int u = q[st++];
        for(int i = head[u]; ~i; i = e[i].nextt) {
            int v = e[i].to;
            if(!dis[v]&&e[i].c) {
                dis[v] = dis[u]+1;
                if(v == t)
                    return 1;
                q[ed++] = v;
            }
        }
    }
    return 0;
}
int dfs(int u, int limit) {
    if(u == t) {
        return limit;
    }
    int cost = 0;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(e[i].c&&dis[v] == dis[u]+1) {
            int tmp = dfs(v, min(limit-cost, e[i].c));
            if(tmp>0) {
                e[i].c -= tmp;
                e[i^1].c += tmp;
                cost += tmp;
                if(cost == limit)
                    break;
            } else {
                dis[v] = -1;
            }
        }
    }
    return cost;
}
int dinic() {
    int ans = 0;
    while(bfs()) {
        ans += dfs(s, inf);
    }
    return ans;
}
struct edge
{
    int u, v, w;
}ed[maxn];
int main()
{
    int n, m, x;
    int maxx = 0;
    cin>>n>>m>>x;
    for(int i = 0; i<m; i++) {
        scanf("%d%d%d", &ed[i].u, &ed[i].v, &ed[i].w);
        maxx = max(maxx, ed[i].w);
    }
    double l = 0, r = maxx;
    int j = 0;
    while(j<100) {
        double mid = (l+r)/2;
        s = 1, t = n;
        init();
        for(int i = 0; i<m; i++) {
            add(ed[i].u, ed[i].v, int(min(ed[i].w/mid, x*1.0+1e-3)));
        }
        if(dinic() >= x) {
            l = mid;
        } else {
            r = mid;
        }
        j++;
    }
    printf("%.8f\n", l*x);
    return 0;
}