[BZOJ 1733] [Usaco2005 feb] Secret Milking Machine 【二分 + 最大流】

时间:2025-05-01 19:05:49

题目链接:BZOJ - 1733

题目分析

直接二分这个最大边的边权,然后用最大流判断是否可以有 T 的流量。

代码

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm> using namespace std; const int MaxN = 200 + 5, MaxM = 80000 + 5, INF = 999999999; inline int gmin(int a, int b) {return a < b ? a : b;}
inline int gmax(int a, int b) {return a > b ? a : b;} int n, m, Ts, Top, Ans, S, T, Tot;
int d[MaxN], Num[MaxN]; struct Edge
{
int u, v, w, len;
Edge *Next, *Other;
} E[MaxM * 2], *P = E, *Point[MaxN], *Last[MaxN], *EA[MaxM]; inline void AddEdge(int x, int y, int z)
{
Edge *Q = ++P; ++P; EA[++Top] = P;
P -> u = x; P -> v = y; P -> len = z;
P -> Next = Point[x]; Point[x] = P; P -> Other = Q;
Q -> u = y; Q -> v = x; Q -> len = z;
Q -> Next = Point[y]; Point[y] = Q; Q -> Other = P;
} void Set_Edge(int x)
{
for (int i = 1; i <= Top; ++i)
{
if (EA[i] -> len <= x) EA[i] -> w = 1;
else EA[i] -> w = 0;
EA[i] -> Other -> w = 0;
}
} int DFS(int Now, int Flow)
{
if (Now == T) return Flow;
int ret = 0;
for (Edge *j = Last[Now]; j; j = j -> Next)
if (j -> w && d[j -> u] == d[j -> v] + 1)
{
Last[Now] = j;
int p = DFS(j -> v, gmin(j -> w, Flow - ret));
j -> w -= p; j -> Other -> w += p; ret += p;
if (ret == Flow) return ret;
}
if (d[S] >= Tot) return ret;
if (--Num[d[Now]] == 0) d[S] = Tot;
++Num[++d[Now]];
Last[Now] = Point[Now];
return ret;
} bool Check()
{
int MaxFlow = 0;
S = 1; T = n; Tot = n;
memset(d, 0, sizeof(d));
memset(Num, 0, sizeof(Num)); Num[0] = Tot;
for (int i = 1; i <= Tot; ++i) Last[i] = Point[i];
while (d[S] < Tot) MaxFlow += DFS(S, INF);
if (MaxFlow >= Ts) return true;
else return false;
} int main()
{
scanf("%d%d%d", &n, &m, &Ts);
int a, b, c;
int l, r, mid;
l = INF; r = -INF;
Top = 0;
for (int i = 1; i <= m; ++i)
{
scanf("%d%d%d", &a, &b, &c);
AddEdge(a, b, c);
AddEdge(b, a, c);
l = gmin(l, c);
r = gmax(r, c);
}
while (l <= r)
{
mid = (l + r) >> 1;
Set_Edge(mid);
if (Check())
{
Ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
printf("%d\n", Ans);
return 0;
}