POJ 3498 March of the Penguins point flow with limited maximum flow

Test instructions

Give the coordinates of the n block of ice, the number of penguins on each ice pack and the number of times they can withstand the jump, and ask what ice floe will bring the penguins together.

Analysis:

The limit of the point traffic is converted to the limit of the edge traffic, and the maximum flow is calculated.

Code:

POJ 3498//sep9 #include <iostream> #include <queue> #include <algorithm> #include <cmath> usin
G namespace Std;
const int maxn=128;
const int MAXM=40;
const int maxv=260;
const int maxe=26000; struct Edge {int u,v,f,nxt;}
E[MAXE*2+10];
Queue<int> que;
int src,sink;
int g[maxv+10];
int nume;
BOOL VIS[MAXV+10];

int dist[maxv+10];
Double X[MAXN],Y[MAXN];

int NUM[MAXN],M[MAXN];
	void Addedge (int u,int v,int c) {e[nume].u=u,e[nume].v=v;e[nume].f=c;e[nume].nxt=g[u];g[u]=nume++;
e[nume].u=v;e[nume].v=u;e[nume].f=0;e[nume].nxt=g[v];g[v]=nume++;	
	} void Init () {memset (g,0,sizeof (g));
nume=2;
	} 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);
				dist[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;
		} else Dist[e[i].v]=-1;
} return ret;
	} int dinic () {int ret=0;
	while (BFS () ==1) Ret+=dfs (Src,int_max);	
return ret;
	} void Solve () {int i,j,t,n,sum;
	Double D;
	scanf ("%d%lf", &n,&d);
	Sum=0,src=2*n;
	int ok=0;
		for (i=0;i<n;++i) {scanf ("%lf%lf%d%d", &x[i],&y[i],&num[i],&m[i]);
	Sum+=num[i];		
		} for (t=0;t<n;++t) {init ();
			for (i=0;i<n;++i) {Addedge (src,i,num[i]);
		Addedge (I,i+n,m[i]); } for (I=0;i<n;++i) for (j=0;j<n;++j) if (i!=j) {double l=sqrt ((X[i]-x[j]) * (X[i]-x[j]) + (Y[i]-y[j]) * (y[i]-
					Y[J])); if (l<d| |
					Fabs (l-d) <1e-6) {Addedge (I+n,j,int_max);
		}} sink=t; if (Dinic () ==sum) {printf ("%d", t);
		ok=1;
	}} if (!ok) printf ("-1");
printf ("\ n");
	} int main () {int cases;
	scanf ("%d", &cases);
	while (cases--) solve ();	
return 0;   }