zoj 3691 Flower(二分+最大流)

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题意：给出n个点，每个点上有F[i]个花，现在要把所有的花搬到点1，每次最多能走的距离为R，但是可以再中间某个点停下，然后接着走。每个点不能停留超过L[i]次，求R的最小值使得所有的花都能搬到1。

思路：比较明显的二分+网络流吧。二分R的值，然后拆点，点i向i+n连一条容量为L[i]的边。判断两两的点之间是否可达，如果可达，相应的连边，看最大流是否等于F[i]的和。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<stack>#include<set>#include<cmath>#include<vector>#define inf 0x3f3f3f3f#define Inf 0x3FFFFFFFFFFFFFFFLL#define eps 1e-9#define pi acos(-1.0)using namespace std;typedef long long ll;const int maxn=200+50;const int maxm=200000+10;struct Edge{    int to,cap,flow,next;    Edge(){}    Edge(int to,int cap,int flow,int next):to(to),cap(cap),flow(flow),next(next){}}edges[maxm<<1];int head[maxn],cur[maxn],d[maxn],nEdge,n;void AddEdges(int from,int to,int cap){    edges[++nEdge]=Edge(to,cap,0,head[from]);    head[from]=nEdge;    edges[++nEdge]=Edge(from,0,0,head[to]);    head[to]=nEdge;}bool BFS(int s,int t){    memset(d,0xff,sizeof(d));    queue<int>q;    q.push(s);    d[s]=0;    while(!q.empty())    {        int u=q.front();q.pop();        for(int k=head[u];k!=-1;k=edges[k].next)        {            Edge e=edges[k];            if(d[e.to]==-1&&e.cap>e.flow)            {                d[e.to]=d[u]+1;                q.push(e.to);            }        }    }    return d[t]!=-1;}int DFS(int u,int a,int t){    if(u==t||a==0) return a;    int flow=0,f;    for(int &k=cur[u];k!=-1;k=edges[k].next)    {        Edge e=edges[k];        if(d[e.to]==d[u]+1&&(f=DFS(e.to,min(a,e.cap-e.flow),t))>0)        {            edges[k].flow+=f;            edges[k^1].flow-=f;            flow+=f;a-=f;            if(a==0) break;        }    }    return flow;}int MaxFlow(int s,int t){    int flow=0;    while(BFS(s,t))    {        for(int i=0;i<=n*2+1;++i) cur[i]=head[i];        flow+=DFS(s,inf,t);    }    return flow;}struct Point{    double x,y,z;    void read()    {        scanf("%lf%lf%lf",&x,&y,&z);    }}pt[maxn];int F[maxn],L[maxn];double Dis(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));}int dcmp(double x){    if(fabs(x)<eps) return 0;    return x<0?-1:1;}bool test(double R){    memset(head,0xff,sizeof(head));    nEdge=-1;    int S=n*2+1,T=1,sum=0;    double tmp;    for(int i=2;i<=n;++i)    {        AddEdges(i,i+n,L[i]);        AddEdges(S,i,F[i]);        sum+=F[i];        tmp=Dis(pt[1],pt[i]);        if(dcmp(R-tmp)>=0)            AddEdges(i+n,T,inf);    }    for(int i=2;i<=n;++i)        for(int j=i+1;j<=n;++j)        {            tmp=Dis(pt[i],pt[j]);            if(dcmp(R-tmp)>=0)            {                AddEdges(i+n,j,inf);                AddEdges(j+n,i,inf);            }        }    int ans=MaxFlow(S,T);    return ans==sum;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while(~scanf("%d",&n))    {        for(int i=1;i<=n;++i)        {            pt[i].read();            scanf("%d%d",&F[i],&L[i]);        }        double L=0,R=40000,m;        if(!test(R))            printf("-1\n");        else        {            while(R-L>1e-8)            {                m=(L+R)/2;                if(test(m))                    R=m;                else                    L=m;            }            printf("%.7lf\n",L);        }    }    return 0;}

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