[POJ1637]混合图的欧拉回路判定|网络流

　　Sightseeing tour

Description

　　The city executive board in Lund wants to construct a sightseeing tour by bus in Lund, so that tourists can see every corner of the beautiful city. They want to construct the tour so that every street in the city is visited exactly once. The bus should also start and end at the same junction. As in any city, the streets are either one-way or two-way, traffic rules that must be obeyed by the tour bus. Help the executive board and determine if it's possible to construct a sightseeing tour under these constraints.

program poj1637;
const maxn=210;maxm=80010;
var fa,next,w,rec:array[-1..maxm]of longint;
opt,link,dis,a,b:array[-1..maxn]of longint;
t,test,n,m,ans,j,s,tt,i:longint;
e:array[-1..maxm]of record x,y,z:longint;end;
c:array[-1..maxn,-1..maxn]of longint;

function min(a,b:longint):longint;
begin
if a<b then exit(a) else exit(b);
end;

procedure add(x,y,z:longint);
begin
inc(j);fa[j]:=y;next[j]:=link[x];link[x]:=j;w[j]:=z;rec[j]:=j+1;
inc(j);fa[j]:=x;next[j]:=link[y];link[y]:=j;w[j]:=0;rec[j]:=j-1;
end;

function spfa:boolean;
var head,tail,x,j:longint;
begin
fillchar(dis,sizeof(dis),63);
head:=0;tail:=1;opt[1]:=s;dis[s]:=0;
while head<>tail do
begin
head:=(head+1)mod maxn;
x:=opt[head];j:=link[x];
while j<>0 do
begin
if (dis[x]+1<dis[fa[j]])and(w[j]>0) then
begin
dis[fa[j]]:=dis[x]+1;
tail:=(tail+1) mod maxn;opt[tail]:=fa[j];
end;
j:=next[j];
end;
end;
if dis[t]<>dis[-1] then exit(true) else exit(false);
end;

function dfs(p,sum:longint):longint;
var j,tem,x:longint;
begin
tem:=0;
if p=t then exit(sum);
j:=link[p];
while j<>0 do
begin
if (dis[fa[j]]=dis[p]+1)and(w[j]>0) then
begin
x:=dfs(fa[j],min(sum-tem,w[j]));
inc(tem,x);dec(w[j],x);inc(w[rec[j]],x);
if tem=sum then exit(sum);
end;
j:=next[j];
end;
exit(tem);
end;

function check:boolean;
var i,k:longint;
begin
j:=0;s:=0;t:=n+1;ans:=0;
fillchar(c,sizeof(c),0);
for i:=1 to n do if odd(abs(a[i]-b[i])) then exit(false);
for i:=1 to m do if e[i].z=0 then inc(c[e[i].x,e[i].y]);
for i:=1 to n do if a[i]-b[i]>0 then
begin
add(s,i,(a[i]-b[i]) >> 1);
inc(ans,(a[i]-b[i]) >> 1);
end
else
if b[i]-a[i]>0 then add(i,t,(b[i]-a[i]) >> 1);
for i:=1 to n do
for k:=1 to n do if c[i,k]>0 then add(i,k,c[i,k]);
while (spfa)and(ans>0) do
dec(ans,dfs(s,ans));
if ans=0 then exit(true) else exit(false);
end;

begin
readln(test);
for tt:=1 to test do
begin
readln(n,m);
fillchar(link,sizeof(link),0);
fillchar(next,sizeof(next),0);
for i:=1 to n do
begin
a[i]:=0;b[i]:=0;
end;
for i:=1 to m do
begin
readln(e[i].x,e[i].y,e[i].z);
inc(a[e[i].x]);inc(b[e[i].y]);
end;
if check then writeln('possible') else writeln('impossible');
end;
end.