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\n",L);
107     return 0;
108 }
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posted @ 2017-09-18 13:55  Blue233333  阅读(378)  评论(0)  编辑  收藏  举报