P1401 City (30 points, positive solution network stream), p140130 points

P1401 City (30 points, positive solution network stream), p140130 points
Description

N (2 <= n <= 200) cities, M (1 <= m <= 40000) undirected edges, you need to find a T (1 <= T <= 200) route from City 1 to city N, which minimizes the length of the longest side and the edge cannot be reused.

Input/Output Format

Input Format:

 

The three integers N, M, and T in the 1st rows are separated by spaces.

Rows 2nd to P + 1. Each line contains three integers: Ai and Bi. Li indicates that there is a road with a length of Li between city Ai and city Bi.

 

Output Format:

 

The output contains only one row and an integer, that is, the minimum length of the longest path passing through these roads.

 

Input and Output sample input sample #1:

7 9 21 2 22 3 53 7 51 4 14 3 14 5 75 7 11 6 36 7 3

Output sample #1:

5

The positive solution is network flow...
So it's embarrassing ..
But the second answer is still written.
When I get online, I must be back with this question ..

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 using namespace std; 7 const int MAXN=40001; 8 void read(int & n) 9 {10     char c='+';int x=0;bool flag=0;11     while(c<'0'||c>'9')12     {c=getchar();if(c=='-')flag=1;}13     while(c>='0'&&c<='9')14     {x=x*10+(c-48);c=getchar();}15     flag==1?n=-x:n=x;16 }17 int n,m,t;18 struct node19 {20     int u,v,w,nxt,use;21 }edge[MAXN];22 int head[MAXN];23 int num=1;24 void add_edge(int x,int y,int z)25 {26     edge[num].u=x;27     edge[num].v=y;28     edge[num].w=z;29     edge[num].nxt=head[x];30     head[x]=num++;31 }32 int maxl=-1,minl=0x7fff;33 int vis[MAXN];34 int map[201][201];35 int have[201][201];36 int bfs(int need)37 {38     queue<int>q;39     q.push(1);40     while(q.size()!=0)41     {42         int p=q.front();43         q.pop();44         for(int i=head[p];i!=-1;i=edge[i].nxt)45         {46             if(edge[i].w<=need&&have[edge[i].u][edge[i].v]==0)47             {48                 have[edge[i].u][edge[i].v]=1;49                 have[edge[i].v][edge[i].u]=1;50                 if(edge[i].v!=n)51                 q.push(edge[i].v);52                 vis[edge[i].v]++;53             }54         }    55     }56     if(vis[n]>=t)57         return 1;58     else 59         return 0;60     61 }62 int pd(int p)63 {64     memset(vis,0,sizeof(vis));65     memset(have,0,sizeof(have));66     if(bfs(p))67         return 1;68     else 69         return 0;70 }71 int main()72 {73     read(n);read(m);read(t);74     for(int i=1;i<=n;i++)75         head[i]=-1;76     for(int i=1;i<=m;i++)77     {78         int x,y,z;79         read(x);read(y);read(z);80         add_edge(x,y,z);81         add_edge(y,x,z);82         maxl=max(maxl,z);83         minl=min(minl,z);84     }85     int l=minl,r=maxl;86     while(l<r)87     {88         int mid=(l+r)>>1;89         if(pd(mid))90             r=mid;91         else l++;92     }93     printf("%d",l);94     return 0;95 }