IndiaHacks 2016 D. Delivery Bears(二分、最大流)

题意:

$给定N\le 50个城市,M\le 500条有向边,X\le10^5为熊的个数$
$边描述为(u_i,v_i,c_i),表示u_i\to v_i可以通过c_i物品$
$现要求恰好用X只熊,且每只熊运送的物品多少相同$
$求最多能从1到n运多少物品$

分析:

$首先很显然要二分每只熊运送多少物品,答案就是X*这个值$
$接下来如何check可行,显然可以转换一下模型$
$把每条边运送物品变为在当前的mid物品下,可以通过多少只熊$
$所以check就变成了,1到n能否至少通过X只熊,建图后判断最大流是不是\ge X$
$需要注意的是,容量可能会爆int$

代码:

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//
// Created by TaoSama on 2016-03-19
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 50 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 1000 + 10;
int head[N], pnt[M], cap[M], nxt[M], cnt;
void add_edge(int u, int v, int w) {
pnt[cnt] = v;
cap[cnt] = w;
nxt[cnt] = head[u];
head[u] = cnt++;
}
void add_double(int u, int v, int w1, int w2 = 0) {
add_edge(u, v, w1);
add_edge(v, u, w2);
}
int lev[N], cur[N];
bool bfs(int s, int t) {
queue<int> q;
memset(lev, 0, sizeof lev);
q.push(s); lev[s] = 1;
while(q.size() && !lev[t]) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(cap[i] > 0 && !lev[v]) {
lev[v] = lev[u] + 1;
q.push(v);
}
}
}
return lev[t];
}
int dfs(int u, int t, int delta) {
if(u == t || !delta) return delta;
int ret = 0;
for(int i = cur[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(cap[i] > 0 && lev[v] == lev[u] + 1) {
int d = dfs(v, t, min(delta, cap[i]));
cur[u] = i;
ret += d; delta -= d;
cap[i] -= d;
cap[i ^ 1] += d;
if(delta == 0) return ret;
}
}
lev[u] = 0;
return ret;
}
int dinic(int s, int t) {
int ret = 0;
while(bfs(s, t)) {
for(int i = s; i <= t; ++i) cur[i] = head[i];
ret += dfs(s, t, INF);
}
return ret;
}
int n, m, k;
int u[M], v[M], c[M];
bool check(double x) {
cnt = 0; memset(head, -1, sizeof head);
for(int i = 1; i <= m; ++i)
add_double(u[i], v[i], min(c[i] / x, 1e9));
return dinic(1, n) >= k;
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
scanf("%d%d%d", &n, &m, &k);
for(int i = 1; i <= m; ++i) scanf("%d%d%d", u + i, v + i, c + i);
double l = 0, r = 1e6;
for(int i = 1; i <= 100; ++i) {
double mid = (l + r) / 2;
if(check(mid)) l = mid;
else r = mid;
}
printf("%.12f\n", l * k);
return 0;
}


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