题目描述

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.

输入

* 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.

输出

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

样例输入

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

样例输出

5

题解

二分+网络流最大流

显然最大长度满足二分性质，我们可以二分长度mid，这样只能选择长度小于等于mid的边来走。

然后由于题目限制了每条边只能走一次，要求1到n的路径数，显然这是一个最大流问题。

对于原图中的边x->y，长度为z，如果z<=mid，则连边x->y，容量为1；否则不连边。

然后跑最大流，判断是否大于等于T即可，并对应调整上下界。

说实话这道网络流真是再水不过了。

#include <cstdio>#include <cstring>#include <queue>#define N 210#define M 40010using namespace std;queue<int> q;int n , m , p , x[M] , y[M] , z[M] , head[N] , to[M << 2] , val[M << 2] , next[M << 2] , cnt , s , t , dis[N];void add(int x , int y , int z){	to[++cnt] = y , val[cnt] = z , next[cnt] = head[x] , head[x] = cnt;	to[++cnt] = x , val[cnt] = 0 , next[cnt] = head[y] , head[y] = cnt;}bool bfs(){	int x , i;	memset(dis , 0 , sizeof(dis));	while(!q.empty()) q.pop();	dis[s] = 1 , q.push(s);	while(!q.empty())	{		x = q.front() , q.pop();		for(i = head[x] ; i ; i = next[i])		{			if(val[i] && !dis[to[i]])			{				dis[to[i]] = dis[x] + 1;				if(to[i] == t) return 1;				q.push(to[i]);			}		}	}	return 0;}int dinic(int x , int low){	if(x == t) return low;	int temp = low , i , k;	for(i = head[x] ; i ; i = next[i])	{		if(val[i] && dis[to[i]] == dis[x] + 1)		{			k = dinic(to[i] , min(temp , val[i]));			if(!k) dis[to[i]] = 0;			val[i] -= k , val[i ^ 1] += k;			if(!(temp -= k)) break;		}	}	return low - temp;}bool judge(int mid){	int i , ans = 0;	memset(head , 0 , sizeof(head)) , cnt = 1;	for(i = 1 ; i <= m ; i ++ )		if(z[i] <= mid)			add(x[i] , y[i] , 1) , add(y[i] , x[i] , 1);	while(bfs()) ans += dinic(s , 0x7fffffff);	return ans >= p;}int main(){	int i , l = 0 , r = 0 , mid , ans = -1;	scanf("%d%d%d" , &n , &m , &p) , s = 1 , t = n;	for(i = 1 ; i <= m ; i ++ ) scanf("%d%d%d" , &x[i] , &y[i] , &z[i]) , r = max(r , z[i]);	while(l <= r)	{		mid = (l + r) >> 1;		if(judge(mid)) ans = mid , r = mid - 1;		else l = mid + 1;	}	printf("%d\n" , ans);	return 0;}

【bzoj1733】[Usaco2005 feb]Secret Milking Machine 神秘的挤奶机 二分+网络流最大流