Secret Milking Machine POJ - 2455 二分最大流

Farmer John is constructing a new milking machine and wishes to keep it secret as long as possible. He has hidden in it deep within his farm and needs to be able to get to the machine without being detected. He must make a total of T (1 <= T <= 200) trips to the machine during its construction. He has a secret tunnel that he uses only for the return trips. 

The farm comprises N (2 <= N <= 200) landmarks (numbered 1..N) connected by P (1 <= P <= 40,000) bidirectional trails (numbered 1..P) and with a positive length that does not exceed 1,000,000. Multiple trails might join a pair of landmarks. 

To minimize his chances of detection, FJ knows he cannot use any trail on the farm more than once and that he should try to use the shortest trails. 

Help FJ get from the barn (landmark 1) to the secret milking machine (landmark N) a total of T times. Find the minimum possible length of the longest single trail that he will have to use, subject to the constraint that he use no trail more than once. (Note well: The goal is to minimize the length of the longest trail, not the sum of the trail lengths.) 

It is guaranteed that FJ can make all T trips without reusing a trail.

Input

* Line 1: Three space-separated integers: N, P, and T 

* Lines 2..P+1: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, indicating that a trail connects landmark A_i to landmark B_i with length L_i.

Output

* Line 1: A single integer that is the minimum possible length of the longest segment of Farmer John's route.

Sample Input

7 9 2
1 2 2
2 3 5
3 7 5
1 4 1
4 3 1
4 5 7
5 7 1
1 6 3
6 7 3

Sample Output

5

Hint

Farmer John can travel trails 1 - 2 - 3 - 7 and 1 - 6 - 7. None of the trails travelled exceeds 5 units in length. It is impossible for Farmer John to travel from 1 to 7 twice without using at least one trail of length 5. 

Huge input data,scanf is recommended.

由于每条边只能走一次所以当边流量为1时，汇点最大值便是路径数。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;


const int MAX_D=200+5;
const int MAX_B=400000+5;
const int INF=0x3f3f3f3f;


int tot;

int p[40005][3];

int head[MAX_D];
int iter[MAX_D];
int level[MAX_D];
struct node{
    int to;
    int cap;
    int rev;
    int next;
};
node edge[2*MAX_B];




void add_edge(int a,int b,int c){


    edge[tot].to=b;
    edge[tot].cap=c;
    edge[tot].rev=tot+1;
    edge[tot].next=head[a];
    head[a]=tot++;


    edge[tot].to=a;
    edge[tot].cap=c;
    edge[tot].rev=tot-1;
    edge[tot].next=head[b];
    head[b]=tot++;


}




void bfs(int s){
    memset(level,-1,sizeof(level));
    queue<int> que;
    level[s]=0;
    que.push(s);
    while(!que.empty()){
        int v=que.front();que.pop();
        for(int i=head[v];i!=0;i=edge[i].next){
            node &e=edge[i];
            if(e.cap>0&&level[e.to]<0){
                level[e.to]=level[v]+1;
                que.push(e.to);
            }
        }
    }
}


int dfs(int v,int t,int f){
    if(v==t)return f;
    for(int &i=iter[v];i!=0;i=edge[i].next){
        node &e = edge[i];
        if(e.cap>0&&level[v]<level[e.to]){
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0){
                e.cap-=d;
                edge[e.rev].cap+=d;
                return d;
            }
        }
    }


    return 0;
}


int max_flow(int s,int t){
    int flow=0;
    while(1){
        bfs(s);
        if(level[t]<0)return flow;
        for(int i=0;i<MAX_D;i++){
            iter[i]=head[i];
        }
        int f;
        while((f=dfs(s,t,INF))>0){
            flow+=f;
        }
    }
}




void solve(){
    int N,P,T;
    while(~scanf("%d%d%d",&N,&P,&T)){
        int mi=INF;
        int ma=0;
        for(int i=1;i<=P;i++){
            scanf("%d%d%d",&p[i][0],&p[i][1],&p[i][2]);
            mi=min(mi,p[i][2]);
            ma=max(ma,p[i][2]);
        }

        int left=mi,right=ma,ans=0;

        while(left<=right){
            int mid=(left+right)>>1;
            tot=1;
            memset(head,0,sizeof(head));
            for(int i=1;i<=P;i++){
                if(p[i][2]<=mid){
                    add_edge(p[i][0],p[i][1],1);
                }
            }

            if(max_flow(1,N)>=T){
                ans=mid;
                right=mid-1;
            }
            else{
                left=mid+1;
            }
        }

        printf("%d\n",ans);



    }

}


int main()
{
    solve();


    return 0;
}