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.