poj 1634 Prison rearrangement

Description

In order to lower the risk of riots and escape attempts, the boards of two nearby prisons of equal prisoner capacity, have decided to rearrange their prisoners among themselves. They want to exchange half of the prisoners of one prison, for half of the prisoners of the other. However, from the archived information of the prisoners' crime history, they know that some pairs of prisoners are dangerous to keep in the same prison, and that is why they are separated today, i.e. for every such pair of prisoners, one prisoners serves time in the first prison, and the other in the second one. The boards agree on the importance of keeping these pairs split between the prisons, which makes their rearrangement task a bit tricky. In fact, they soon find out that sometimes it is impossible to fulfil their wish of swapping half of the prisoners. Whenever this is the case, they have to settle for exchanging as close to one half of the prisoners as possible.

Input

On the first line of the input is a single positive integer n, telling the number of test scenarios to follow. Each scenario begins with a line containing two non-negative integers m and r, 1 < m < 200 being the number of prisoners in each of the two prisons, and r the number of dangerous pairs among the prisoners. Then follow r lines each containing a pair xi yi of integers in the range 1 to m,which means that prisoner xi of the first prison must not be placed in the same prison as prisoner yi of the second prison.

Output

For each test scenario, output one line containing the largest integer k <= m/2 , such that it is possible to exchange k prisoners of the first prison for k prisoners of the second prison without getting two prisoners of any dangerous pair in the same prison.

 

題目大意：有兩個監獄都有n人，要分別抽調不多於n div 2的人交換。原先兩個監獄之間有些犯人對必須在不同的監獄。給出e對犯人不能同在一個監獄。求最大交換數。

//===========================================================================

把給出的每對犯人連雙向邊，求出聯通塊，容易發現一個聯通塊裏要換一起換。求有幾個聯通塊，做01揹包。輸出f[max,max]。

嗯。這裏註明一下。今天WA了將近十次，因爲循環變量應該從大到小for，不然一個塊會取很多遍。。。揹包都編挫了。。。。。。。。。。。。。。

還有，這題內存限制很小。。。大家自重。。。。

 

AC CODE

 

program pku_1636;
var map:array[1..400,0..200] of longint;
    p:array[1..400] of boolean;
    w:array[0..1,1..400] of longint;
    f:array[0..100,0..100] of boolean;
    i,tot,len,size,n:longint;
//============================================================================
procedure dfs(u:longint);
var v,i:longint;
begin p[u]:=true;
  if u>n then inc(w[1,size]) else
    inc(w[0,size]);
  for i:=1 to map[u,0] do
    if not(p[map[u,i]]) then
      dfs(map[u,i]);
end;
//============================================================================
procedure work;
var m,i,j,k,x,y:longint;
begin
  readln(n,m);
  fillchar(map,sizeof(map),0);
  for i:=1 to m do
  begin
    readln(x,y); inc(y,n);
    inc(map[x,0]); map[x,map[x,0]]:=y;
    inc(map[y,0]); map[y,map[y,0]]:=x;
  end; size:=0;
  fillchar(p,sizeof(p),0);
  for i:=1 to n*2 do
    if not(p[i]) then
    begin
      inc(size); w[0,size]:=0;
      w[1,size]:=0; dfs(i);
    end;
  fillchar(f,sizeof(f),0);
  f[0,0]:=true; n:=n div 2;
  for k:=1 to size do
    for i:=n downto w[0,k] do
      for j:=n downto w[1,k] do
        f[i,j]:=f[i-w[0,k],j-w[1,k]] or f[i,j];
  for i:=n downto 0 do
    if f[i,i] then
    begin
      writeln(i);
      exit;
    end;
end;
//============================================================================
begin
  readln(tot);
  for i:=1 to tot do work;
end.