网络流24题之最长k可重线段集问题

对于每个线段拆成两个点，如同之前一样建图，由于可能出现垂直于x轴的

所以建图由i指向i~

继续最小费用最大流

By：大奕哥

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=10000005,inf=1e9;
 4 int head[N],d[N],f[N],l1[N],r1[N],l2[N],r2[N],a[N],s=1e9,t,n,k,cnt=-1;
 5 long long cost;
 6 bool v[N];
 7 struct node{
 8     int to,nex,f,w,c;
 9 }e[1000005];
10 void add(int x,int y,int w,int c)
11 {
12     e[++cnt].to=y;e[cnt].w=w;e[cnt].f=x;e[cnt].c=c;e[cnt].nex=head[x];head[x]=cnt;
13     e[++cnt].to=x;e[cnt].w=0;e[cnt].f=y;e[cnt].c=-c;e[cnt].nex=head[y];head[y]=cnt;
14 }
15 queue<int>q;
16 bool spfa()
17 {
18     memset(f,-1,sizeof(f));
19     memset(d,0x3f,sizeof(d));
20     memset(v,0,sizeof(v));
21     d[s]=0;v[s]=1;q.push(s);
22     while(!q.empty())
23     {
24         int x=q.front();q.pop();v[x]=0;
25         for(int i=head[x];i!=-1;i=e[i].nex)
26         {
27             int y=e[i].to;
28             if(d[y]<=d[x]+e[i].c||!e[i].w)continue;
29
30             d[y]=d[x]+e[i].c;f[y]=i;
31             if(!v[y])q.push(y),v[y]=1;
32         }
33     }
34     if(d[t]>1e9)return 0;
35     int flow=inf;
36     for(int i=f[t];i!=-1;i=f[e[i].f])
37     flow=min(flow,e[i].w);
38     for(int i=f[t];i!=-1;i=f[e[i].f])
39     e[i].w-=flow,e[i^1].w+=flow,cost+=1ll*e[i].c*flow;
40     return 1;
41 }
42 int main()
43 {
44     scanf("%d%d",&n,&k);int num=0;
45     memset(head,-1,sizeof(head));
46     for(int i=1;i<=n;++i)
47     {
48         scanf("%d%d%d%d",&l1[i],&r1[i],&l2[i],&r2[i]);
49         a[++num]=l1[i];a[++num]=l2[i];
50     }
51     sort(a+1,a+1+num);
52     num=unique(a+1,a+1+num)-a-1;
53     for(int i=1;i<=n;++i)
54     {
55         int x=sqrt(1ll*(l1[i]-l2[i])*(l1[i]-l2[i])+1ll*(r2[i]-r1[i])*(r2[i]-r1[i]));
56         l1[i]=lower_bound(a+1,a+1+num,l1[i])-a;
57         l2[i]=lower_bound(a+1,a+1+num,l2[i])-a;
58         if(l1[i]!=l2[i])
59         add((l1[i]<<1)|1,l2[i]<<1,1,-x);
60         else
61         add(l1[i]<<1,(l2[i]<<1)|1,1,-x);
62     }
63     for(int i=1;i<num;++i)
64     {
65         add((i<<1)|1,i+1<<1,inf,0);
66         add(i<<1,(i<<1)|1,inf,0);
67     }
68     add(num<<1,(num<<1)|1,inf,0);
69     t=num*2+10;
70     add((num<<1)|1,t,k,0);
71     add(0,2,k,0);s=0;
72     while(spfa());
73     printf("%lld\n",-cost);
74     return 0;
75 }

原文地址：https://www.cnblogs.com/nbwzyzngyl/p/8400477.html