ZOJ 3691 Flower 解题报告

比赛

题意：给出n个点的三维坐标，每个点上有f盆花，每个点至多只能运出l盆花（有些花是从别的点运到这个点的），只有距离小于r的点可以相互运输，问r最少为多少，可以让所有点的花都运到第一个点上。

解法：最大流。把一个点拆成两个aa‘，连容量为l的边，从原点向a连容量为f的边，距离小于r的两点连容量为inf的边。由点1向汇点连边。

//Memory: 3824 KB
//Time: 280 MS
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define MAXN 405
#define MAXM 200005
#define INF 1000000007
#define EPS 1e-8
struct Edge{
int v,w,next;
}edge[MAXM];
struct Point{
double x,y,z;
int f,l;
void input(){scanf("%lf%lf%lf%d%d",&x,&y,&z,&f,&l);}
}p[MAXN];
double Dist(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 en,head[MAXN];
int h[MAXN],gap[MAXN];
double dist[MAXN][MAXN];
void init()
{
en = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
edge[en].v = v,edge[en].w = w;
edge[en].next = head[u];
head[u] = en++;
swap(u,v);
edge[en].v = v,edge[en].w = 0;
edge[en].next = head[u];
head[u] = en++;
}
void build(double m,int n)
{
init();
for(int i = 1; i<=n; i++)
{
addedge(0,i,p[i].f);
addedge(i,i+n,p[i].l);
}
addedge(1+n,n*2+1,INF);
for(int i = 2; i <= n; i++)
for(int j = 1; j <= n; j++)
{
if(i==j||dist[i][j]+EPS>m) continue;
addedge(i+n,j,INF);
}

}
int dfs(int s,int t,int pos,int n,int cost)
{
if(pos==t) return cost;
int minh = n-1,lv = cost,d;
for(int j = head[pos]; j!=-1; j= edge[j].next)
{
int v = edge[j].v, w = edge[j].w;
if(w>0)
{
if(h[v]+1==h[pos])
{
d = min(lv,w);
d = dfs(s,t,v,n,d);
edge[j].w-=d,edge[j^1].w+=d;
lv-=d;
if(h[s]>=n) return cost-lv;
if(lv==0) break;
}
if(h[v]<minh) minh = h[v];
}
}
if(lv==cost)
{
gap[h[pos]]--;
if(gap[h[pos]]==0) h[s] = n;
h[pos] = minh+1;
gap[h[pos]]++;
}
return cost - lv;
}
int sap(int st,int ed,int n)
{
int ret = 0;
memset(gap,0,sizeof(gap));
memset(h,0,sizeof(h));
gap[0] = n;
while(h[st]<n) ret+=dfs(st,ed,st,n,INF);
return ret;
}
int main()
{
//freopen("/home/moor/Code/input","r",stdin);
int n,sum;
bool flag;
double l,r,mid,ans;

while(scanf("%d",&n)!=EOF)
{
l = r = 0;
sum = 0;
flag = 1;
for(int i = 1; i <= n; i++)
{
p[i].input();
sum+=p[i].f;
if(i!=1&&p[i].f>p[i].l) flag = 0;
}
if(!flag)
{
printf("-1\n");
continue;
}
for(int i = 1; i <=n; i++)
for(int j = i+1; j<=n; j++)
dist[i][j] = dist[j][i] = Dist(p[i],p[j]),
r = max(r,dist[i][j]);
r+=10;
p[1].l = INF;
ans = -1;
while(r-l>EPS)
{
mid = (l+r)/2;
build(mid,n);
if(sap(0,n*2+1,n*2+2)==sum) ans = mid,r = mid-EPS;
else l=mid+EPS;
}
if(ans<0) printf("-1\n");
else printf("%.8f\n",ans);
}
return 0;
}