POJ 2455 Secret Milking machine two points + maximum flow

Test instructions

For a graph, a path that does not intersect the T-bar Edge from point 1 to N, and the goal is to minimize the maximum edge in the T-path and output the maximum edge.

Analysis:

The problem of finding the most value satisfies monotonicity can be done by two points, and the second is an accelerated enumeration method. The two-point enumeration of the maximum edge build, each time with a length of less than or equal to two points of the edge of the map and the capacity of 1, the maximum flow can be.

Code:

POJ 2455//sep9#include <iostream> #include <queue> #include <algorithm>using namespace Std;const int Maxn=256;const int maxm=40012;struct edge{int v,f,nxt;} e[maxm*2+10];queue<int> que;int src,sink;int g[maxn+10];int nume;bool vis[maxn+10];int dist[maxN+10];int n,p,t; int a[maxm],b[maxm],c[maxm];void addedge (int u,int v,int c) {e[++nume].v=v;e[nume].f=c;e[nume].nxt=g[u];g[u]=nume;e[ ++nume].v=u;e[nume].f=c;e[nume].nxt=g[v];g[v]=nume;} void Init () {memset (g,0,sizeof (g)); nume=1;} int BFs () {while (!que.empty ()) Que.pop (); memset (dist,0,sizeof (Dist)); memset (vis,0,sizeof (Vis)); vis[src]=true; Que.push (SRC), while (!que.empty ()) {int U=que.front (), Que.pop (), for (int i=g[u];i;i=e[i].nxt) if (e[i].f&&! VIS[E[I].V]) {Que.push (E[I].V);d ist[e[i].v]=dist[u]+1;vis[e[i].v]=true; if (e[i].v==sink) return 1;}} return 0;} int dfs (int u,int delta) {if (U==sink) return delta;int ret=0;for (int i=g[u];ret<delta&&i;i=e[i].nxt) if (E[i] . f&&dist[e[i].v]==dist[u]+1) {int Dd=dfs (e[i].v,min(E[i].f,delta-ret)); if (dd>0) {e[i].f-=dd;e[i^1].f+=dd;ret+=dd;} Elsedist[e[i].v]=-1;} return ret;} int dinic () {int ret=0;while (BFS () ==1) Ret+=dfs (Src,int_max); return ret;} BOOL Check (int mid) {init (); src=0,sink=n+1;for (int i=0;i<p;++i) if (c[i]<=mid) Addedge (a[i],b[i],1); Addedge (src , 1,t); Addedge (n,sink,t); if (Dinic () ==t) return True;return false;} int main () {scanf ("%d%d%d", &n,&p,&t), int maxl=-1;for (int i=0;i<p;++i) {scanf ("%d%d%d", &a[i], &b[i],&c[i]); Maxl=max (Maxl,c[i]);} int L=0,r=maxl+1,ans;while (l<r) {int mid= (l+r)/2;if (check (mid)) r=mid,ans=mid;elsel=mid+1;}  printf ("%d", ans); return 0;}

POJ 2455 Secret Milking machine two points + maximum flow