题意：有n块浮冰，每块上有一定数量的企鹅，每块浮冰可以承受企鹅跳一定的次数后消失，给出企鹅跳的最大距离，求所有企鹅能跳到哪块浮冰上见面。
建图：添加源点，汇点，源点与浮冰相连，容量为浮冰上企鹅的数量，浮冰之间可以相互连边，所以要拆点，拆完后点与拆点间连边，容量为浮冰能承受企鹅跳的次数，如果两块浮冰i,j间可以到达，加边i+n—>j,j+n—>i,容量无穷大。然后枚举所有浮冰与汇点相连求最大流。
#include<stdio.h>
#include<string.h>
const int N=210;
const int inf=0x3fffffff;
int dis[N],gap[N],head[N],num,start,end,ans,n;
struct edge
{
int st,ed,flow,next;
}e[N*N];
struct node
{
double x,y;
int cont,po;
}P[210];
void addedge(int x,int y,int w)
{
e[num].st=x;e[num].ed=y;e[num].flow=w;e[num].next=head[x];head[x]=num++;
e[num].st=y;e[num].ed=x;e[num].flow=0;e[num].next=head[y];head[y]=num++;
}
double Dis(int i,int j)
{
return (P[i].x-P[j].x)*(P[i].x-P[j].x)+(P[i].y-P[j].y)*(P[i].y-P[j].y);
}
int dfs(int u,int minflow)
{
if(u==end)return minflow;
int i,v,f,flow=0,min_dis=ans-1;
for(i=head[u];i!=-1;i=e[i].next)
{
if(e[i].flow>0)
{
v=e[i].ed;
if(dis[v]+1==dis[u])
{
f=dfs(v,e[i].flow>minflow-flow?minflow-flow:e[i].flow);
e[i].flow-=f;
e[i^1].flow+=f;
flow+=f;
if(flow==minflow)break;
if(dis[start]>=ans)return flow;
}
min_dis=min_dis>dis[v]?dis[v]:min_dis;
}
}
if(flow==0)
{
if(--gap[dis[u]]==0)
dis[start]=ans;
dis[u]=min_dis+1;
gap[dis[u]]++;
}
return flow;
}
int isap()
{
int maxflow=0;
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
gap[0]=ans;
while(dis[start]<ans)
maxflow+=dfs(start,inf);
return maxflow;
}
int main()
{
int i,j,sum,t,flag,k;
double D;
scanf("%d",&t);
while(t--)
{
scanf("%d%lf",&n,&D);
flag=0;sum=0;
start=n+n;end=n+n+1;ans=end+1;
for(i=0;i<n;i++)
{
scanf("%lf%lf%d%d",&P[i].x,&P[i].y,&P[i].cont,&P[i].po);
sum+=P[i].cont;
}
for(k=0;k<n;k++)
{
memset(head,-1,sizeof(head));num=0;
addedge(k,end,inf);
for(i=0;i<n;i++)
{
addedge(i,i+n,P[i].po);
addedge(start,i,P[i].cont);
for(j=i+1;j<n;j++)
{
if(Dis(i,j)<=D*D)
{
addedge(i+n,j,inf);
addedge(j+n,i,inf);
}
}
}
if(isap()==sum)
{
printf("%d ",k);
flag++;
}
}
if(flag==0)printf("-1\n");
else printf("\n");
}
return 0;
}