POJ2455 Secret Milking Machine "two points, maximum flow"

The main idea: N points P-bar edge, so that there is a path from 1 to N of the T bar, the maximum value of the Benquan on the path to the minimum number

Idea: Do a lot of two points + the maximum flow of the problem, the idea is very good out of two the largest side of the right to build a map, run Dinic

The problem is .... This problem is a good question of the card constant!!!!!

T after 8 hair really can't stand, aimed at the eyes of the online program, Uniform adjacency matrix .... On the superiority of adjacency matrix

But not superstitious I finally repeated the optimization constants after the adjacency table a

Program for Tle

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <queue>

#define MAXN 200090

#define ESP 0.001

#define INF 0x3f3f3f3f

using namespace Std;

int head[300],next[maxn],point[maxn],now=0;

int flow[maxn],dist[300];

int tt,p,h=0,n;

struct T

{

int X;int Y;int v;

}A[MAXN];

void Add (int x,int Y,int v)

{

NEXT[++NOW]=HEAD[X];

Head[x]=now;

Point[now]=y;

Flow[now]=v;

Next[++now]=head[y];

Head[y]=now;

Point[now]=x;

flow[now]=0;

}

int BFS (int s,int t,int x)

{

queue<int>q;

Q.push (s);

memset (dist,-1,sizeof (Dist));

dist[s]=0;

while (!q.empty ())

{

int U=q.front ();

Q.pop ();

for (int i=head[u];i;i=next[i])

{

int k=point[i];

if (flow[i]!=0&&dist[k]==-1)

{

dist[k]=dist[u]+1;

Q.push (k);

}

}

}

return dist[t]!=-1;

}

int dfs (int s,int d,int t,int x)

{

if (s==t) return D;

int res=0;

for (int i=head[s];i&&res<d;i=next[i])

{

int u=point[i];

if (flow[i]&&dist[u]==dist[s]+1)

{

int Dd=dfs (U,min (flow[i],d-res), t,x);

if (DD)

{

FLOW[I]-=DD;

flow[((i-1) ^1) +1]+=dd;

RES+=DD;

}

}

}

if (res==0) dist[s]=-1;

return res;

}

int judge (int x,int s,int t)

{

int ans=0;

memset (head,0,sizeof (head));

now=0;

for (int i=1;i<=p;i++) if (a[i].v<=x)

{

Add (a[i].x,a[i].y,1);

Add (a[i].y,a[i].x,1);

}

Add (S,1,tt); add (N,t,inf);

while (BFS (s,t,x))

{Ans+=dfs (s,inf,t,x);}

if (ANS>=TT) return 1;else return 0;

}

int read ()

{

int X=0,f=1;char Ch=getchar ();

while (ch< ' 0 ' | | Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}

while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();}

return x*f;

}

int main ()

{

int x,y,v;

int l=0x3f3f3f3f,r=0,mid;

scanf ("%d%d%d", &N,&P,&TT);

int s=n+10,t=n+12;

for (int i=1;i<=p;i++)

{

X=read (); Y=read (); V=read ();

A[i].x=x;a[i].y=y;a[i].v=v;

R=max (R,V);

L=min (L,V);

}

while (mid= (l+r) >>1,l<r)

{

if (judge (mid,s,t) ==1) R=mid;else l=mid+1;

}

printf ("%d\n", R);

return 0;

}

The AC program

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <queue>

#define MAXN 200090

#define ESP 0.001

#define INF 0x3f3f3f3f

using namespace Std;

int head[300],next[maxn],point[maxn],now=0;

int flow[maxn],dist[300];

int tt,p,h=0,n;

struct T

{

int X;int Y;int v;

}A[MAXN];

void Add (int x,int Y,int v)

{

NEXT[++NOW]=HEAD[X];

Head[x]=now;

Point[now]=y;

Flow[now]=v;

Next[++now]=head[y];

Head[y]=now;

Point[now]=x;

flow[now]=0;

}

int BFS (int s,int t,int x)

{

queue<int>q;

Q.push (s);

memset (dist,-1,sizeof (Dist));

dist[s]=0;

while (!q.empty ())

{

int U=q.front ();

Q.pop ();

for (int i=head[u];i;i=next[i])

{

int k=point[i];

if (flow[i]!=0&&dist[k]==-1)

{

dist[k]=dist[u]+1;

Q.push (k);

}

}

}

return dist[t]!=-1;

}

int dfs (int s,int d,int t,int x)

{

if (s==t) return D;

int res=0;

for (int i=head[s];i&&res<d;i=next[i])

{

int u=point[i];

if (flow[i]&&dist[u]==dist[s]+1)

{

int Dd=dfs (U,min (flow[i],d-res), t,x);

if (DD)

{

FLOW[I]-=DD;

flow[((i-1) ^1) +1]+=dd;

RES+=DD;

}

}

}

if (res==0) dist[s]=-1;

return res;

}

int judge (int x,int s,int t)

{

int ans=0;

memset (head,0,sizeof (head));

now=0;

for (int i=1;i<=p;i++) if (a[i].v<=x)

{

Add (a[i].x,a[i].y,1);

Add (a[i].y,a[i].x,1);

}

Add (S,1,tt); add (N,t,inf);

while (BFS (s,t,x))

{Ans+=dfs (s,inf,t,x);}

if (ANS>=TT) return 1;else return 0;

}

int read ()

{

int X=0,f=1;char Ch=getchar ();

while (ch< ' 0 ' | | Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}

while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();}

return x*f;

}

int main ()

{

int x,y,v;

int l=0x3f3f3f3f,r=0,mid;

scanf ("%d%d%d", &N,&P,&TT);

int s=n+10,t=n+12;

for (int i=1;i<=p;i++)

{

X=read (); Y=read (); V=read ();

A[i].x=x;a[i].y=y;a[i].v=v;

R=max (R,V);

L=min (L,V);

}

while (mid= (l+r) >>1,l<r)

{

if (judge (mid,s,t) ==1) R=mid;else l=mid+1;

}

printf ("%d\n", R);

return 0;

}

Finish is also drunk, not a don't sleep tut

POJ2455 Secret Milking Machine "two points, maximum flow"