UVA12125 March of the Penguins (最大流+拆点)

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;
}