POJ 2135 Farm Tour minimum fee inflow door template

Test instructions

Ask for point 1 to N and then from Point N to point 1 without the shortest path of the same route.

Analysis:

It is easy to build a set of data for the shortest two times:

4 5
1 2 1
1 3 100
2 4 100
2 3 1
3 4 1

The next minimum fee flow template is good.

Code:

POJ 2135//sep9 #include <iostream> #include <queue> using namespace std;
const int maxn=2048;
const int maxm=20024;	
struct Edge {int v,f,w,nxt;
}E[4*MAXM+10];
int g[maxn+10];
int nume,src,sink;
Queue<int> Q;
BOOL INQ[MAXN+10];
int dist[maxn+10];

int prev[maxn+10],pree[maxn+10];

int n,m;	
	void Addedge (int u,int v,int c,int w) {e[++nume].v=v;e[nume].f=c;e[nume].w=w;e[nume].nxt=g[u];g[u]=nume;
E[++nume].v=u;e[nume].f=0;e[nume].w=-w;e[nume].nxt=g[v];g[v]=nume; } bool Find_path () {while (!
	Q.empty ()) Q.pop ();	
	memset (dist,0x7f,sizeof (Dist));
	memset (inq,false,sizeof (INQ));
	Q.push (SRC);
	Inq[src]=true;
	dist[src]=0; while (! Q.empty ()) {int U=q.front ();
		Q.pop ();
		Inq[u]=false; for (int i=g[u];i;i=e[i].nxt) {if (e[i].f>0&&dist[u]+e[i].w<dist[e[i].v]) {dist[e[i].v]=dist[u]+e[i].
				W
				Prev[e[i].v]=u;	
				Pree[e[i].v]=i;
					if (!INQ[E[I].V]) {Q.push (E[I].V);
				Inq[e[i].v]=true; }}}} return dist[sink]<dist[maxn]?trUe:false;
	} int Min_cost_flow (int f) {int res=0;	
		while (f>0) {if (Find_path () ==false) return-1;
		int d=f;
		for (int v=sink;v!=src;v=prev[v]) d=min (D,E[PREE[V]].F);
		F-=d;
		Res+=d*dist[sink];
			for (int v=sink;v!=src;v=prev[v]) {e[pree[v]].f-=d;
		E[pree[v]^1].f+=d;			
}} return res;
	} void Init () {memset (g,0,sizeof (g));
nume=1;
	} int main () {scanf ("%d%d", &n,&m);
	Init ();
		while (m--) {int a,b,c;
		scanf ("%d%d%d", &a,&b,&c);
		Addedge (A-1,B-1,1,C);
	Addedge (B-1,A-1,1,C);
	} src=0,sink=n-1;
	printf ("%d\n", Min_cost_flow (2));	
return 0;  }