zoukankan      html  css  js  c++  java
  • poj——1273 Drainage Ditches(费用流模版)

    Description

    Every time it rains on Farmer John’s fields, a pond forms over Bessie’s favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie’s clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
    Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
    Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
    Input

    The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
    Output

    For each case, output a single integer, the maximum rate at which water may emptied from the pond.

    中文题意:点击出翻译

    题解:

    简单来说就是费用流,这里是用johnson算法。
    

    代码:

    const
      maxn=40001;
      maxm=201;
    type
      arr=record
            x,y,c,w,next:longint;
          end;
    var
      n,m,nm,vs,vt:longint;
      a:array [-1..maxn] of arr;
      ls,dis,q:array [0..maxm] of longint;
      v:array [0..maxm] of boolean;
    function min(o,p:longint):longint;
    begin
      if o<p then exit(o);
      exit(p);
    end;
    
    procedure add(x,y,c:longint);
    begin
      inc(nm);
      a[nm].x:=x; a[nm].y:=y; a[nm].c:=c;
      a[nm].next:=ls[x]; ls[x]:=nm;
    end;
    
    function bfs:boolean;
    var
      head,tail,xx,i:longint;
    begin
      fillchar(v,sizeof(v),false);
      tail:=1; head:=0; dis[1]:=1;
      v[1]:=true; q[1]:=1;
      while head<tail do
        begin
          inc(head); xx:=q[head];
          i:=ls[xx];
          while i<>-1 do
            with a[i] do
              begin
                if (not v[y]) and (w<c) then
                  begin
                    v[y]:=true;
                    dis[y]:=dis[xx]+1;
                    inc(tail);
                    q[tail]:=y;
                  end;
                i:=next;
              end;
        end;
      exit(v[vt]);
    end;
    
    function addflow(p,maxflow:longint):longint;
    var
      i,o:longint;
    begin
      if (p=vt) or (maxflow=0) then exit(maxflow);
      addflow:=0; i:=ls[p];
      while i<>-1 do
        with a[i] do
          begin
            if (dis[y]=dis[p]+1) and (w<c) then
              begin
                o:=addflow(y,min(maxflow,c-w));
                if o>0 then
                  begin
                    inc(w,o);
                    dec(a[i xor 1].w,o);
                    dec(maxflow,o);
                    inc(addflow,o);
                    if maxflow=0 then break;
                  end;
              end;
            i:=next;
          end;
    end;
    
    procedure init;
    var
      i,x,y,c:longint;
    begin
      fillchar(ls,sizeof(ls),255);
      fillchar(a,sizeof(a),0);
      readln(n,m); nm:=0;
      for i:=1 to n do
         begin
           readln(x,y,c);
           add(x,y,c);
           add(y,x,0);
         end;
      vs:=1; vt:=m;
    end;
    
    procedure main;
    var
      ans:longint;
    begin
      ans:=0;
      while bfs do
        ans:=ans+addflow(vs,maxlongint);
      writeln(ans);
    end;
    
    begin
      while not eof do
        begin
          init;
          main;
        end;
    end.
    
    
  • 相关阅读:
    每次运行caffe代码之前需要考虑修改的地方
    caffe solver 配置详解
    python获取当前文件路径以及父文件路径
    Python 文件夹及文件操作
    安装NVIDIA驱动时禁用自带nouveau驱动
    博客园转载其他博客园的文章:图片和源码
    分布式开放消息系统(RocketMQ)的原理与实践
    RocketMQ基本概念及原理介绍
    rocketmq 4.3.2 解决远程不能消费问题,解决未识别到公网IP问题
    osx免驱网卡推荐
  • 原文地址:https://www.cnblogs.com/zyx-crying/p/9319507.html
Copyright © 2011-2022 走看看