• BZOJ1733: [Usaco2005 feb]Secret Milking Machine 神秘的挤奶机


    n<=200个点m<=40000条边无向图,求   t次走不经过同条边的路径从1到n的经过的边的最大值   的最小值。

    最大值最小--二分,t次不重边路径--边权1的最大流。

      1 #include<stdio.h>
      2 #include<string.h>
      3 #include<stdlib.h>
      4 #include<algorithm>
      5 //#include<iostream>
      6 using namespace std;
      7 
      8 int n,m,t;
      9 #define maxn 211
     10 #define maxm 160011
     11 struct Edge{int to,next,v;}edge[maxm];int first[maxn],le=2;
     12 void in(int x,int y,int v) {Edge &e=edge[le];e.to=y;e.v=v;e.next=first[x];first[x]=le++;}
     13 void insert(int x,int y,int v) {in(x,y,v);in(y,x,v);}
     14 const int inf=0x3f3f3f3f;
     15 struct network
     16 {
     17     struct Edge{int to,next,cap,flow;}edge[maxm];int first[maxn],le,n,s,t;
     18     void clear(int n)
     19     {
     20         memset(first,0,sizeof(first));
     21         le=2;this->n=n;
     22     }
     23     void in(int x,int y,int cap) {Edge &e=edge[le];e.to=y;e.cap=cap;e.flow=0;e.next=first[x];first[x]=le++;}
     24     void insert(int x,int y,int cap) {in(x,y,cap);in(y,x,0);}
     25     int que[maxn],head,tail,dis[maxn],cur[maxn];
     26     bool bfs()
     27     {
     28         memset(dis,0,sizeof(dis));
     29         dis[s]=1;
     30         que[head=(tail=1)-1]=s;
     31         while (head!=tail)
     32         {
     33             const int now=que[head++];
     34             for (int i=first[now];i;i=edge[i].next)
     35             {
     36                 const Edge &e=edge[i];
     37                 if (e.cap>e.flow && !dis[e.to])
     38                 {
     39                     dis[e.to]=dis[now]+1;
     40                     que[tail++]=e.to;
     41                 }
     42             }
     43         }
     44         return dis[t];
     45     }
     46     int dfs(int x,int a)
     47     {
     48         if (x==t || !a) return a;
     49         int flow=0,f;
     50         for (int &i=cur[x];i;i=edge[i].next)
     51         {
     52             Edge &e=edge[i];
     53             if (dis[e.to]==dis[x]+1 && (f=dfs(e.to,min(a,e.cap-e.flow)))>0)
     54             {
     55                 flow+=f;
     56                 e.flow+=f;
     57                 edge[i^1].flow-=f;
     58                 a-=f;
     59                 if (!a) break;
     60             }
     61         }
     62         return flow;
     63     }
     64     int dinic(int s,int t)
     65     {
     66         this->s=s;this->t=t;
     67         int ans=0;
     68         while (bfs())
     69         {
     70             for (int i=1;i<=n;i++) cur[i]=first[i];
     71             ans+=dfs(s,inf);
     72         }
     73         return ans;
     74     }
     75 }g;
     76 bool check(int x)
     77 {
     78     g.clear(n);
     79     for (int i=1;i<=n;i++)
     80         for (int j=first[i];j;j=edge[j].next)
     81         {
     82             const Edge &e=edge[j];
     83             if (e.v<=x) g.insert(i,e.to,1);
     84         }
     85     return g.dinic(1,n)>=t;
     86 }
     87 int x,y,v;
     88 int main()
     89 {
     90     scanf("%d%d%d",&n,&m,&t);
     91     int Max=0;
     92     memset(first,0,sizeof(first));
     93     for (int i=1;i<=m;i++)
     94     {
     95         scanf("%d%d%d",&x,&y,&v);
     96         insert(x,y,v);
     97         Max=max(Max,v);
     98     }
     99     int L=0,R=Max+1;
    100     while (L<R)
    101     {
    102         int mid=(L+R)>>1;
    103         if (check(mid)) R=mid;
    104         else L=mid+1;
    105     }
    106     printf("%d
    ",L);
    107     return 0;
    108 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Blue233333/p/7542758.html
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