Hihocoder Network Stream One · Ford-fulkerson algorithm

Network Flow One · Ford-fulkerson algorithm time limit: 10000ms single point time limit: 1000ms memory limit: 256MB description

Small hi and Small ho live in P city, p City is a big big city, so also faced with a big city will encounter problems: traffic congestion.

Little ho: Every weekend comes home feeling traffic jam is a kind of suffering ah.

Small hi: Usually the traffic is also good, just one to the rush hour will be more crowded.

Little ho: If you can limit the number of cars, I do not know if there is a way to know the maximum traffic flow system, so that can be limited to a can always be a smooth quantity.

Little hi: Theoretically, there are algorithms. As early as 1955, T.E Harris raised the question of seeking maximum traffic between two points on a given network. and thus a new graph theory model is created: network flow.

Little ho: What exactly is that?

Little hi: The mathematical description of the language is given a graph g= (v,e), where each edge (U,V) has a non-negative capacity value, recorded as C (U,v) ≥0. At the same time there are two special vertices in the graph, the source point S and the meeting point T.

As an example:

Where node 1 is the source point S and Node 6 is the meeting point T.

We ask for the maximum feasible flow from the source point S to the meeting point T, which is also known as the maximum flow problem.

In this example, the maximum flow is 5, respectively: 1→2→4→6, Flow is 1;1→3→4→6, flow is 2;1→3→5→6, flow is 2.

Little ho: It looks like fun, you let me think about it first.

Hint: Ford-fulkerson algorithm

Input

Line 1th: 2 positive integer n,m. 2≤n≤500,1≤m≤20,000.

2nd.. M+1 line: 3 integers per line u,v,c (U,V), representing an edge (U,V) and its capacity C (u,v). 1≤u,v≤n,0≤c (u,v) ≤100.

The default source point in the given diagram is 1, and the meeting point is N. There may be duplicate edges.

Output

Line 1th: An integer representing the maximum stream for the given figure G.

Sample input

6 71 2 31 3 52 4 13 4 23 5 34 6 45 6 2

Sample output

5
Analysis: The key to the maximum flow is to seek the augmented road, adjacent table storage edge, easy to modify;
Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#include<climits>#include<cstring>#include<string>#include<Set>#include<map>#include<queue>#include<stack>#include<vector>#include<list>#defineRep (I,m,n) for (i=m;i<=n;i++)#defineRSP (It,s) for (Set<int>::iterator It=s.begin (); It!=s.end (); it++)#defineMoD 1000000007#defineINF 0x3f3f3f3f#defineVI vector<int>#definePB Push_back#defineMP Make_pair#defineFi first#defineSe Second#definell Long Long#definePi ACOs (-1.0)#definePII pair<int,int>#defineLson L, Mid, rt<<1#defineRson mid+1, R, rt<<1|1Const intmaxn=5e2+Ten;using namespacestd;ll gcd (ll p,ll q) {returnq==0? P:GCD (q,p%q);} ll Qpow (ll p,ll q) {ll F=1; while(q) {if(q&1) f=f*p;p=p*p;q>>=1;}returnF;}intN,m,k,t,h[maxn],tot,p[maxn],a[maxn],v[maxn][maxn],ans;structnode{intFr,to,nxt,cap,flow;} e[20000<<1];voidAddintXintYintz) {e[tot].fr=x; E[tot].to=y; E[TOT].NXT=H[x]; E[tot].cap=Z; H[X]=tot++; e[tot].fr=y; E[tot].to=x; E[TOT].NXT=H[y]; H[y]=tot++;}voidMax_flow (intSintt) {ans=0;  while(1) {memset (A,0,sizeofa); Queue<int>Q; Q.push (s); A[s]=inf;  while(!Q.empty ()) {            intx=Q.front ();            Q.pop ();  for(inti=h[x];i!=-1; i=e[i].nxt) {                intto=e[i].to,cap=e[i].cap,flow=E[i].flow; if(!a[to]&&cap>flow) {P[to]=i; A[to]=min (a[x],cap-flow);                Q.push (to); }            }            if(A[t]) Break; }        if(!a[t]) Break;  for(intnow=t;now!=s;now=e[p[now]].fr) {E[p[now]].flow+=A[t]; E[p[now]^1].flow-=A[t]; } ans+=A[t]; }}intMain () {inti,j; scanf ("%d%d",&n,&m); memset (H,-1,sizeofh);  while(m--)    {        intb,c,d; scanf ("%d%d%d",&b,&c,&d); V[B][C]+=D; } Rep (I,1, N) Rep (J,1, N)if(V[i][j]) Add (I,j,v[i][j]); Max_flow (1, N); printf ("%d\n", ans); //System ("Pause");    return 0;}

Hihocoder Network Stream One · Ford-fulkerson algorithm