HDU 3549 Basic network flow EK algorithm flow problem

Welcome to the--bestcoder Anniversary (High quality topic + multiple Rewards)
Flow problem

Time limit:5000/5000 MS (java/others) Memory limit:65535/32768 K (java/others)
Total submission (s): 10184 Accepted Submission (s): 4798

Problem Descriptionnetwork Flow is a well-known difficult problem for acmers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

Inputthe first line of input contains an integer T, denoting the number of the test cases.
For each test case, the first line contains integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <=, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is a edge from X to Y and the capacity of it is C. (1 & lt;= X, Y <= N, 1 <= C <= 1000)

Outputfor Each test cases, you should output the maximum flow from source 1 to sink N.

Sample Input23 21 2 12 3 13 31 2 12 3 11 3 1

Sample outputcase 1:1case 2:2

Authorhyperhexagon

Source Hyperhexagon ' s Summer Gift (Original tasks)

Recommendzhengfeng | We have carefully selected several similar problems for you:1532 3572 3416 3081 3491

#include <stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<queue>using namespacestd;intdp[ -][ -],pre[ -];Const inttmin=999999999;intMaxflow;voidEK (intStartintEndintN) {     while(1) {Queue<int>Q; Q.push (1); intminflow=tmin; memset (PRE,0,sizeof(pre));  while(!Q.empty ()) {            intu=Q.front ();            Q.pop ();  for(intI=1; i<=n;i++){                if(dp[u][i]>0&&!Pre[i]) {Pre[i]=u;                Q.push (i); }            }        }        if(pre[end]==0)             Break;  for(intI=end;i!=start;i=Pre[i]) {Minflow=min (dp[pre[i]][i],minflow); }         for(intI=end;i!=start;i=Pre[i]) {Dp[pre[i]][i]-=Minflow; Dp[i][pre[i]]+=Minflow; } Maxflow+=Minflow; }}intMain () {intCount=0; intn,m; intT; scanf ("%d",&6);  while(t--) {scanf ("%d%d",&n,&m); Memset (DP,0,sizeof(DP)); memset (PRE,0,sizeof(pre)); Count++; intu,v,w;  for(intI=1; i<=m;i++) {scanf ("%d%d%d",&u,&v,&W); DP[U][V]+=W; } Maxflow=0; EK (1, N,n); printf ("Case %d:%d\n", Count,maxflow); }   return 0;}

HDU 3549 Basic network flow EK algorithm flow problem