题意:
有N个点,每个点有ni个企鹅,这个点最多能跳着离开mi次.企鹅每次最多能跳D单位远,每两点之间有坐标距离..要你求出哪些点可以聚集所有的企鹅
分析:
因为每个点最多离开mi次,说明这个点有结点容量(允许通过的最大容量),故拆点i*2,i*2+1;详细见代码
// File Name: 12125.cpp // Author: Zlbing // Created Time: 2013/4/23 20:42:02 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack> #include<cmath> #include<queue> using namespace std; #define CL(x,v); memset(x,v,sizeof(x)); #define INF 0x3f3f3f3f #define LL long long #define REP(i,r,n) for(int i=r;i<=n;i++) #define RREP(i,n,r) for(int i=n;i>=r;i--) const int MAXN=200+100; struct Edge{ int from,to,cap,flow; }; bool cmp(const Edge& a,const Edge& b){ return a.from < b.from || (a.from == b.from && a.to < b.to); } struct node{ int x,y,n,m; }P[MAXN]; struct Dinic{ int n,m,s,t; vector<Edge> edges; vector<int> G[MAXN]; bool vis[MAXN]; int d[MAXN]; int cur[MAXN]; void init(int n){ this->n=n; for(int i=0;i<=n;i++)G[i].clear(); edges.clear(); } void AddEdge(int from,int to,int cap){ edges.push_back((Edge){from,to,cap,0}); edges.push_back((Edge){to,from,0,0});//当是无向图时，反向边容量也是cap,有向边时，反向边容量是0 m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BFS(){ CL(vis,0); queue<int> Q; Q.push(s); d[s]=0; vis[s]=1; while(!Q.empty()){ int x=Q.front(); Q.pop(); for(int i=0;i<G[x].size();i++){ Edge& e=edges[G[x][i]]; if(!vis[e.to]&&e.cap>e.flow){ vis[e.to]=1; d[e.to]=d[x]+1; Q.push(e.to); } } } return vis[t]; } int DFS(int x,int a){ if(x==t||a==0)return a; int flow=0,f; for(int& i=cur[x];i<G[x].size();i++){ Edge& e=edges[G[x][i]]; if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){ e.flow+=f; edges[G[x][i]^1].flow-=f; flow+=f; a-=f; if(a==0)break; } } return flow; } //当所求流量大于need时就退出，降低时间 int Maxflow(int s,int t,int need){ this->s=s;this->t=t; int flow=0; while(BFS()){ CL(cur,0); flow+=DFS(s,INF); if(flow>need)return flow; } return flow; } //最小割割边 vector<int> Mincut(){ BFS(); vector<int> ans; for(int i=0;i<edges.size();i++){ Edge& e=edges[i]; if(vis[e.from]&&!vis[e.to]&&e.cap>0)ans.push_back(i); } return ans; } void Reduce(){ for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow; } void ClearFlow(){ for(int i = 0; i < edges.size(); i++) edges[i].flow = 0; } }; Dinic solver; int main() { int cas; scanf("%d",&cas); while(cas--) { int N; double D; scanf("%d%lf",&N,&D); int sum=0; REP(i,1,N){ scanf("%d%d%d%d",&P[i].x,&P[i].y,&P[i].n,&P[i].m); sum+=P[i].n; } solver.init(2*N+1); int s=0; int t; REP(j,1,N) { solver.AddEdge(j*2,j*2+1,P[j].m); solver.AddEdge(s,j*2,P[j].n); } REP(j,1,N) REP(k,j+1,N) { if(sqrt((P[j].x-P[k].x)*(P[j].x-P[k].x)+(P[j].y-P[k].y)*(P[j].y-P[k].y))<=D) { solver.AddEdge(j*2+1,k*2,INF); solver.AddEdge(k*2+1,j*2,INF); } } vector<int> ans; REP(i,1,N) { t=i*2; solver.ClearFlow(); int ret=solver.Maxflow(s,t,INF); if(ret==sum) ans.push_back(i-1); } if(ans.size()) { for(int i=0;i<ans.size();i++) { if(i==0)printf("%d",ans[i]); else printf(" %d",ans[i]); } printf("\n"); } else printf("-1\n"); } return 0; }