POJ2455 Secret Milking Machine ——二分答案+网络流——Pku2455

思路：二分答案，用权值比二分出来的答案小的边来建图。网络流判定的是是否能够满足找到t条路径。

CODE

Program Secret;//By_Poetshy   
Const  
    maxn=40000;  
Var  
    i,j,k,m,n,p,t,d         :Longint;  
    pre,other,last,data     :Array[1..maxn*2]of Longint;  
    a,b                     :Array[1..200]of Longint;  
    v                       :Array[1..200]of Boolean;  
    f,c                     :Array[1..200,1..200]of Longint;  
    l,r,mid,ans,max         :Longint;  
  
Function Min(i,j:Longint):Longint;  
begin  
    if i<j then exit(i);exit(j);  
end;  
  
Function Check(i:Longint):Boolean;  
var j,k,p,q,sum,head,tail   :Longint;  
    seq                     :Array[1..maxn]of Longint;  
begin  
    fillchar(f,sizeof(f),0);  
    fillchar(c,sizeof(c),0);  
    for p:=1 to n do  
        begin  
            j:=last[p];  
            while j<>0 do  
                begin  
                    k:=other[j];  
                    if data[j]<=i then inc(c[p,other[j]]);  
                    j:=pre[j];  
                end;  
        end;  
    repeat  
        fillchar(v,sizeof(v),0);  
        head:=0;tail:=1;seq[1]:=1;  
        a[1]:=0;b[1]:=maxlongint;  
        v[1]:=true;  
        while head<tail do  
            begin  
                inc(head);p:=seq[head];  
                for q:=1 to n do  
                    if (c[p,q]<>0)and(f[p,q]<c[p,q])and(not v[q])then  
                        begin  
                            v[q]:=true;  
                            a[q]:=p;b[q]:=Min(c[p,q]-f[p,q],b[p]);  
                            inc(tail);seq[tail]:=q;  
                        end;  
                if seq[tail]=n then break;  
            end;  
        if v[n] then  
            begin  
                p:=n;  
                while p<>0 do  
                    begin  
                        if a[p]<>0 then inc(f[a[p],p],b[n]);  
                        p:=a[p];  
                    end;  
            end;  
    until not v[n];  
    sum:=0;  
    for p:=1 to n-1 do inc(sum,f[p,n]);  
    if sum>=t then exit(true);  
    exit(false);  
end;  
  
BEGIN  
    readln(n,p,t);  
    for i:=1 to p do  
        begin  
            readln(m,j,d);  
            inc(k);pre[k]:=last[m];last[m]:=k;other[k]:=j;data[k]:=d;  
            inc(k);pre[k]:=last[j];last[j]:=k;other[k]:=m;data[k]:=d;  
            if d>max then max:=d;  
        end;  
    l:=0;r:=max;  
    while l<=r do  
        begin  
            mid:=(l+r)>>1;  
            if check(mid) then  
                begin  
                    ans:=mid;  
                    r:=mid-1;  
                end else l:=mid+1;  
        end;  
    writeln(ans);  
END.