# POJ 2455 Secret Milking Machine "two points + maximum flow" "1 to n non-repeating path not less than T bar, the minimum value of the maximum edge weight value on the selected path is obtained"

Source: Internet
Author: User

Secret Milking Machine
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10620 Accepted: 3110

Description

Farmer John is constructing a new milking, and wishes to keep it secret as long as possible. He has hidden in it deep within his farm and needs to is able to get to the machine without being detected. He must make a total of T (1 <= t <=) 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 <=) landmarks (numbered 1..N) connected by P (1 <= P <= 40,000) bidirectional Trails (numbered 1..P) and with a positive length this does not exceed 1,000,000. Multiple trails might join a pair of landmarks.

To minimize him chances of detection, FJ knows he cannot use any trail in the farm more than once and that he should try T o 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'll have to use, subject to the constraint that he u Se No trail more than once. (Note well:the goal is to minimize the length of the longest trail and not the sum of the trail lengths.)

It is guaranteed this 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 land Mark A_i to landmark b_i with length l_i.

Output

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

Sample Input

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

Long time No 1 a ...

Test Instructions: gives you an n-point (numbered from 1 to N) and M-bars without a graph and weights for each edge. Requires at least a path from 1 to n that does not duplicate the T-bar, allowing you to find the minimum value of the maximum edge value on all paths (of course, the chosen ones, not the chosen ones), in order to satisfy this premise. the topic guarantees that from 1 to n there will be at least one path that is not duplicated on the T-bar, i.e. the problem must be solved.

Idea: Two-minute enumeration of the minimum value of the maximum edge weight on all selected paths, mid, to determine if there are at least one path where the T-edge is not duplicated at the limit of the mid value from 1 to N.

The number of paths that are not duplicated by the edge can be solved with the maximum flow-enumerating all the non-forward edges in the original, and if the edge is not greater than mid, the edge is added to the new graph and the edge capacity is 1. Finally, run the maximum stream from 1 to N.

AC Code:

`#include <cstdio> #include <cstring> #include <queue> #include <stack> #include <vector># Include <algorithm> #define MAXN 200+10#define maxm 1600000+10#define INF 0x3f3f3f3fusing namespace Std;struct Edge {int from, to, cap, flow, next;}; Edge EDGE[MAXM], Redge[maxm];int HEAD[MAXN], Edgenum, RHEAD[MAXN], Redgenum;int DIST[MAXN], Cur[maxn];bool Vis[MAXN]; int N, M, t;struct node{int from, to, Val, next;};    Node Node[maxm];int HEAD[MAXN], nodenum;void addNode (int u, int v, int w) {node E = {u, V, W, Head[u]};    Node[nodenum] = E; Head[u] = nodenum++;}    int max;//record maximum edge right void input () {memset (head,-1, sizeof (head));    Nodenum = 0;    Max = 0;    int A, b, C;        for (int i = 1; I <= M; i++) {scanf ("%d%d%d", &a, &b, &c);        max = max (max, c);//Update AddNode (A, B, c);    AddNode (b, A, c);    }}void init () {edgenum = 0; Memset (Head,-1, sizeof (head));} void Addedge (int u, int v, int w) {Edge E1 = {u, V, W, 0, Head[u]};    Edge[edgenum] = E1;    Head[u] = edgenum++;    Edge E2 = {V, u, 0, 0, Head[v]};    Edge[edgenum] = E2; HEAD[V] = edgenum++;} void Getmap (int mid) {for (int i = 0; i < Nodenum; i++) {if (Node[i].val <= mid)//less than or equal to limit add    Edge (Node[i].from, node[i].to, 1);//Add New figure}}bool BFS (int s, int t) {queue<int> Q;    memset (Dist,-1, sizeof (Dist));    Memset (Vis, false, sizeof (VIS));    Dist[s] = 0;    Vis[s] = true;    Q.push (s); while (!        Q.empty ()) {int u = q.front ();        Q.pop ();            for (int i = head[u]; i =-1; i = Edge[i].next) {Edge E = Edge[i];                if (!vis[e.to] && e.cap > E.flow) {dist[e.to] = Dist[u] + 1;                Vis[e.to] = true;                if (e.to = = t) return true;            Q.push (e.to); }}} return false;}    int DFS (int x, int a, int t) {if (x = = T | | a = = 0) return A;    int flow = 0, F; for (int &i = CUR[X]; I! =-1;        i = Edge[i].next) {Edge &e = Edge[i]; if (dist[e.to] = = Dist[x] + 1 && (f = DFS (e.to, Min (A, e.cap-e.flow), T)) > 0) {edge[i].flow            + = f;            Edge[i^1].flow-= f;            Flow + + F;            A-= f;        if (a = = 0) break; }} return flow;}    int Maxflow (int s, int t) {int flow = 0;        while (BFS (s, t)) {memcpy (cur, head, sizeof (head));    Flow + = DFS (s, INF, T); } return flow;}        int main () {while (scanf ("%d%d%d", &n, &m, &t)! = EOF) {input ();        int left = 0, right = max, ans = max;            while (right >= left)//binary Find {int mid = (left + right) >> 1;            Init ();            Getmap (mid);  if (Maxflow (1, N) >= T)//Determine if there is a path for T-bar not repeating {ans = min (ans, mid);//Update right = Mid            -1;        } else left = mid + 1; } printf ("%d\n", ans); } return 0;}`

POJ 2455 Secret Milking Machine "two points + maximum flow" "1 to n non-repeating path not less than T bar, the minimum value of the maximum edge weight value on the selected path is obtained"

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.