poj1273

Drainage Ditches

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 61571 Accepted: 23621

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.

Sample Input

5 41 2 401 4 202 4 202 3 303 4 10

Sample Output

50

题意：现在有m个池塘(从1到m开始编号,1为源点,m为汇点),及n条水渠,给出这n条水渠所连接的点和所能流过的最大流量，求从源点到汇点能流过的最大流量。（网络流模板题）


代码：


type
  node=record
    y,r,next,op:longint;
  end;
var
  g:array [1..400] of node;
  h,level,q:array [1..400] of longint;
  i,j,m,n,ans,a,b,c,vs,vt,tot:longint;




function min(x,y:longint):longint;
begin
  if x<y then
  exit(x);
  exit(y);
end;




function dfs(v,a:longint):longint;
var
  tmp,u,ans,value,flow:longint;
begin
  if (v=vt) or (a=0) then exit(a);
  ans:=0;
  tmp:=h[v];
  while tmp<>-1 do
  begin
    u:=g[tmp].y;
    value:=g[tmp].r;
    if level[v]+1=level[u] then
    begin
      flow:=dfs(u,min(a,value));
      if flow<>0 then
      begin
        g[tmp].r:=g[tmp].r-flow;
        g[g[tmp].op].r:=g[g[tmp].op].r+flow;
        ans:=ans+flow;
        a:=a-flow;
        if a=0 then break;
      end;
    end;
    tmp:=g[tmp].next;
  end;
  exit(ans);
end;




function bfs:boolean;
var
  u,tmp,l,f,v:longint;
begin
  fillchar(level,sizeof(level),0);
  q[1]:=vs;
  level[vs]:=1;
  l:=1;
  f:=1;
  repeat
    v:=q[l];
    tmp:=h[v];
    while tmp<>-1 do
    begin
      u:=g[tmp].y;
      if (g[tmp].r<>0) and (level[u]=0) then
      begin
        level[u]:=level[v]+1;
        inc(f);
        q[f]:=u;
        if u=vt then exit(true);
      end;
      tmp:=g[tmp].next;
    end;
    inc(l);
  until l>f;
  exit(false);
end;




procedure add(a,b,c:longint);
begin
  inc(tot);
  g[tot].y:=b;
  g[tot].r:=c;
  g[tot].next:=h[a];
  h[a]:=tot;
  g[tot].op:=tot+1;
  inc(tot);
  g[tot].y:=a;
  g[tot].r:=0;
  g[tot].next:=h[b];
  h[b]:=tot;
  g[tot].op:=tot-1;
end;




begin
  while not eof do
  begin
    fillchar(g,sizeof(g),0);
    fillchar(h,sizeof(h),$ff);
    fillchar(level,sizeof(level),0);
    fillchar(q,sizeof(q),0);
    readln(n,m);
    tot:=0;
    ans:=0;
    vs:=1;
    vt:=m;
    for i:=1 to n do
    begin
      readln(a,b,c);
      add(a,b,c);
    end;
    while bfs do
    ans:=ans+dfs(vs,maxlongint);
    writeln(ans);
  end;
end.