POJ 2230

POJ 2230

题目大意：给出一副联通无向图，求从起点1开始，每条边走两次，回到起点的的一种路径方案。{SPJ}

解：我感觉和欧拉回路有点像，然后是分析性质，把每条无向边拆成两条，可以发现，从任意点的欧拉回路都是存在的（奇数乘偶数=偶数>.<）, 然后我是用了一种无赖走法，把剩余的点看成一个大点，然后走去这个大点再走回来，进入大点后不断缩小规模。主要的问题是判断点是否走过（度走完）以及边是否走过。

//poj 2230
const
        maxn=11111;
        maxm=51111;
type
        data=record
          dest, next, op, cost: longint;
        end;
var
        edge: array[1..maxm*2]of data;
        vect, deg: array[1..maxn]of longint;
        n, m, tot: longint;
procedure add(x, y: longint);
begin
  inc(tot);
  with edge[tot] do begin
    dest := y;
    next := vect[x];
    cost := 1;
    vect[x] := tot;
    op := tot+1;
  end;
  inc(tot);
  with edge[tot] do begin
    dest := x;
    cost := 1;
    next := vect[y];
    vect[y] := tot;
    op := tot-1;
  end;
end;

procedure init;
var
        i, j, x, y: longint;
begin
  fillchar(vect, sizeof(vect), 0);
  fillchar(deg, sizeof(deg), 0);
  tot := 0;
  readln(n, m);
  for i := 1 to m do begin
    readln(x, y);
    add(x, y);
    inc(deg[x], 2); inc(deg[y], 2);
  end;
end;

procedure search(x: longint);
var
        i, u: longint;
begin
  i := vect[x];
  while i<>0 do
    with edge[i] do begin
      if (deg[dest]>0)and(deg[x]>0)and(cost>0) then begin
        dec(deg[x], 2);
        dec(deg[dest], 2);
        cost := 0;
        edge[op].cost := 0;
        writeln(dest);
        search(dest);
        writeln(x);
      end;
      i := next;
    end;
end;

procedure main;
begin
  writeln(1);
  search(1);
end;

begin
  assign(input,'1.txt'); reset(input);
  init;
  main;
end.