• 【POJ3680】Intervals(费用流)


    题意:有n条线段,每条有起点,终点和一个权值

    要求选取一些线段,使它们的权值和最大,并且使每一个点被覆盖不超过k次

    1 ≤ K ≤ N ≤ 200

    1 ≤ ai < bi ≤ 100,000, 1 ≤ wi ≤ 100,000

    思路:RYZ作业

    费用流(经典?)模型之一

    离散化后对于线段(a[i],b[i],w[i]),从a[i]到b[i]连容量为1,费用为w[i]的边

    (i,i+1)之间都连容量为K,费用为0的边,以达到限制max<=k的效果

    S——>1和N——>T之间都连容量为K,费用为0的边

    跑最大费用最大流

      1 var head,vet,next,len1,len2,a,b,c,d,fan:array[1..100000]of longint;
      2     inq:array[1..30000]of boolean;
      3     q:array[0..30000]of longint;
      4     pre:array[1..30000,1..2]of longint;
      5     dis:array[1..30000]of longint;
      6     n,m,k,up,i,tot,ans,s,source,src,cas,v,x,y:longint;
      7 
      8 procedure swap(var x,y:longint);
      9 var t:longint;
     10 begin
     11  t:=x; x:=y; y:=t;
     12 end;
     13 
     14 procedure qsort(l,r:longint);
     15 var i,j,mid:longint;
     16 begin
     17  i:=l; j:=r; mid:=d[(l+r)>>1];
     18  repeat
     19   while mid>d[i] do inc(i);
     20   while mid<d[j] do dec(j);
     21   if i<=j then
     22   begin
     23    swap(d[i],d[j]);
     24    inc(i); dec(j);
     25   end;
     26  until i>j;
     27  if l<j then qsort(l,j);
     28  if i<r then qsort(i,r);
     29 end;
     30 
     31 function min(x,y:longint):longint;
     32 begin
     33  if x<y then exit(x);
     34  exit(y);
     35 end;
     36 
     37 function hash(x:longint):longint;
     38 var l,r,mid:longint;
     39 begin
     40  l:=1; r:=up;
     41  while l<=r do
     42  begin
     43   mid:=(l+r)>>1;
     44   if d[mid]=x then exit(mid);
     45   if d[mid]<x then l:=mid+1
     46    else r:=mid-1;
     47  end;
     48 end;
     49 
     50 procedure add(a,b,c,d:longint);
     51 begin
     52  inc(tot);
     53  next[tot]:=head[a];
     54  vet[tot]:=b;
     55  len1[tot]:=c;
     56  len2[tot]:=d;
     57  head[a]:=tot;
     58 
     59  inc(tot);
     60  next[tot]:=head[b];
     61  vet[tot]:=a;
     62  len1[tot]:=0;
     63  len2[tot]:=-d;
     64  head[b]:=tot;
     65 end;
     66 
     67 function spfa:boolean;
     68 var u,e,v,i,top:longint;
     69 begin
     70  for i:=1 to s do
     71  begin
     72   dis[i]:=-(maxlongint>>1);
     73   inq[i]:=false;
     74  end;
     75  top:=1; q[1]:=source; dis[source]:=0; inq[source]:=true;
     76  while top>0 do
     77  begin
     78   u:=q[top]; dec(top); inq[u]:=false;
     79   e:=head[u];
     80   while e<>0 do
     81   begin
     82    v:=vet[e];
     83    if (len1[e]>0)and(dis[u]+len2[e]>dis[v]) then
     84    begin
     85     dis[v]:=dis[u]+len2[e];
     86     pre[v,1]:=u;
     87     pre[v,2]:=e;
     88     if not inq[v] then
     89     begin
     90      inc(top); q[top]:=v; inq[v]:=true;
     91     end;
     92    end;
     93    e:=next[e];
     94   end;
     95  end;
     96  if dis[src]=-(maxlongint>>1) then exit(false);
     97  exit(true);
     98 end;
     99 
    100 procedure mcf;
    101 var k,e,t:longint;
    102 begin
    103  k:=src; t:=maxlongint;
    104  while k<>source do
    105  begin
    106   t:=min(t,len1[pre[k,2]]);
    107   k:=pre[k,1];
    108  end;
    109  k:=src;
    110  while k<>source do
    111  begin
    112   e:=pre[k,2];
    113   len1[e]:=len1[e]-t;
    114   len1[fan[e]]:=len1[fan[e]]+t;
    115   ans:=ans+t*len2[e];
    116   k:=pre[k,1];
    117  end;
    118 end;
    119 
    120 begin
    121  assign(input,'poj3680.in'); reset(input);
    122  assign(output,'poj3680.out'); rewrite(output);
    123  readln(cas);
    124  for i:=1 to 100000 do
    125   if i and 1=1 then fan[i]:=i+1
    126    else fan[i]:=i-1;
    127  for v:=1 to cas do
    128  begin
    129   for i:=1 to s do head[i]:=0;
    130   s:=0; tot:=0; ans:=0;
    131   //fillchar(head,sizeof(head),0);
    132   read(n,k); m:=0;
    133   for i:=1 to n do
    134   begin
    135    read(a[i],b[i],c[i]);
    136    inc(m); d[m]:=a[i];
    137    inc(m); d[m]:=b[i];
    138   end;
    139   qsort(1,m);
    140   up:=1;
    141   for i:=2 to m do
    142    if d[i]<>d[up] then begin inc(up); d[up]:=d[i]; end;
    143   for i:=1 to n do
    144   begin
    145    x:=hash(a[i]); y:=hash(b[i]);
    146    add(x,y,1,c[i]);
    147   end;
    148   source:=up+1; src:=up+2; s:=up+2;
    149   add(source,1,k,0);
    150   add(up,src,k,0);
    151   for i:=1 to up-1 do add(i,i+1,k,0);
    152   while spfa do mcf;
    153   writeln(ans);
    154  end;
    155  close(input);
    156  close(output);
    157 end.
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  • 原文地址:https://www.cnblogs.com/myx12345/p/6508382.html
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