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[BZOJ1733][Usaco2005 feb]Secret Milking Machine 神秘的挤奶机(二分+最大流)

2017-02-12 19:39 483 查看

题目描述

传送门

题解

这题目就是p

实际上就是问有没有不同的t条路径从1走到n,然后路径的最大值最小

二分显然,二分出mid之后,s->1,T;n->t,T,然后对于每一条边直接连,容量为1

需要注意的是这题是无向边

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
#define N 210
#define E 200005
#define inf 2000000000

int n,m,T,s,t,Max,maxflow,ans;
struct data{int x,y,z;}e[E];
int tot,point
,nxt[E],v[E],remain[E];
int deep
,last
,num
,cur
;
queue <int> q;

void addedge(int x,int y,int cap)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;
}
void bfs(int t)
{
for (int i=1;i<=t;++i) deep[i]=t;
deep[t]=0;
for (int i=1;i<=t;++i) cur[i]=point[i];
while (!q.empty()) q.pop();
q.push(t);
while (!q.empty())
{
int now=q.front();q.pop();
for (int i=point[now];i!=-1;i=nxt[i])
if (deep[v[i]]==t&&remain[i^1])
{
deep[v[i]]=deep[now]+1;
q.push(v[i]);
}
}
}
int addflow(int s,int t)
{
int now=t,ans=inf;
while (now!=s)
{
ans=min(ans,remain[last[now]]);
now=v[last[now]^1];
}
now=t;
while (now!=s)
{
remain[last[now]]-=ans;
remain[last[now]^1]+=ans;
now=v[last[now]^1];
}
return ans;
}
void isap(int s,int t)
{
bfs(t);
for (int i=1;i<=t;++i) ++num[deep[i]];

int now=s;
while (deep[s]<t)
{
if (now==t)
{
maxflow+=addflow(s,t);
now=s;
}

bool has_find=false;
for (int i=cur[now];i!=-1;i=nxt[i])
if (deep[v[i]]+1==deep[now]&&remain[i])
{
has_find=true;
cur[now]=i;
last[v[i]]=i;
now=v[i];
break;
}

if (!has_find)
{
int minn=t-1;
for (int i=point[now];i!=-1;i=nxt[i])
if (remain[i]) minn=min(minn,deep[v[i]]);
if (!(--num[deep[now]])) break;
++num[deep[now]=minn+1];
cur[now]=point[now];
if (now!=s) now=v[last[now]^1];
}
}
}
bool check(int mid)
{
tot=-1;memset(point,-1,sizeof(point));
memset(num,0,sizeof(num));
for (int i=1;i<=m;++i)
if (e[i].z<=mid)
addedge(e[i].x,e[i].y,1),
addedge(e[i].y,e[i].x,1);
addedge(s,1,T),addedge(n,t,T);
maxflow=0;isap(s,t);
return maxflow==T;
}
int find()
{
int l=0,r=Max,mid,ans=Max;
while (l<=r)
{
mid=(l+r)>>1;
if (check(mid)) ans=mid,r=mid-1;
else l=mid+1;
}
return ans;
}
int main()
{
scanf("%d%d%d",&n,&m,&T);
s=n+1,t=s+1;
for (int i=1;i<=m;++i)
{
scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].z);
Max=max(Max,e[i].z);
}
ans=find();
printf("%d\n",ans);
}
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