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 }