二分图的一些题目合集

妈蛋状态都被狗吃了，已经开始不自觉对着电脑发呆……被几道二分图的题亮瞎了双眼（这么弱可是gdkoi只剩一个月gdoi只剩100+天……！！）

 

wikioi1222信与信封问题

二分图，但是判断两个合集中的某两个点是不是只能连在一起。那么我们就在跑一边最大匹配后直接用是否可以增广来判断。如果可以增广那么这两个点是有其他方式连在一起的，否则这两个点就必须连在一起。具体做法是先去掉这两个点的边，不过那么match数组也要跟着改一下。

var
  map:array[0..200,0..200]of boolean;
  matchx,matchy:array[0..200]of longint;
  chose:array[0..200]of boolean;
  n:longint;


function dfs(x:longint):boolean;
var
  i:longint;
begin
  for i:=1 to n do
    if map[x,i] and chose[i] then begin
      chose[i]:=false;
      if (matchx[i]=0) or dfs(matchx[i]) then begin
        matchx[i]:=x;
        matchy[x]:=i;
        exit(true);
      end;
    end;
  exit(false);
end;

procedure into;
var
  j,k:longint;
begin
  readln(n);
  fillchar(map,sizeof(map),true);
  while true do begin
    readln(j,k);
    if j=0 then break;
    map[j,k]:=false;
  end;
end;


function work:boolean;
var
  ans,i,j:longint;
  flag:boolean;
begin
  for i:=1 to n do begin
    fillchar(chose,sizeof(chose),true);
    if not dfs(i) then exit(false);
  end;
  flag:=false;
  for i:=1 to n do begin
    fillchar(chose,sizeof(chose),true);
    j:=matchy[i];
    map[i,j]:=false;
    matchx[j]:=0;
    matchy[i]:=0;
    if not dfs(i) then begin
      writeln(i,' ',j);
      matchx[j]:=i;
      matchy[i]:=j;
      flag:=true;
    end;
    map[i,j]:=true;
  end;
  exit(flag);
end;

begin
  into;
  if not work then writeln('none');
  readln;
end.

 

bzoj2150: 部落战争

最小路径覆盖，等于=点数-最大匹配，把有向图转为二分图的做法就是把每个点分成两个点，一个是入点，如果有边指向这个点则指向入点，一个是出点，如果这个点有指向其他点则出点。这样把原图的边转化一下，然后就直接最大匹配。

type
  arr=record
    toward,next:longint;
  end;

const
  maxm=10000;
  maxn=5000;
var
  edge:array[0..maxm]of arr;
  first,match:array[0..maxn]of longint;
  map:array[0..100,0..100]of boolean;
  chose:array[0..maxn]of boolean;
  n,m,tot,ans:longint;

function check(x,y:longint):boolean;
begin
  exit( (x>=1) and (x<=n) and (y>=1) and (y<=m) and map[x,y] );
end;

function calc(x,y:longint):longint;
begin
  exit((x-1)*m+y);
end;

procedure addedge(j,k:longint);
begin
  inc(tot);
  edge[tot].toward:=k;
  edge[tot].next:=first[j];
  first[j]:=tot;
end;

function dfs(x:longint):boolean;
var
  i,too:longint;
begin
  i:=first[x];
  while i>0 do begin
    too:=edge[i].toward;
    if chose[too] then begin
      chose[too]:=false;
      if (match[too]=0) or dfs(match[too]) then begin
        match[too]:=x;
        exit(true);
      end;
    end;
    i:=edge[i].next;
  end;
  exit(false);
end;

procedure into;
var
  i,j,r,l:longint;
  ch:char;
begin
  readln(n,m,l,r);
  for i:=1 to n do begin
    for j:=1 to m do begin
      read(ch);
      if ch='.' then map[i,j]:=true
        else inc(ans);
    end;
    readln;
  end;
  for i:=1 to n do
    for j:=1 to m do begin
      if check(i+l,j-r) then addedge(calc(i,j),calc(i+l,j-r));
      if check(i+r,j-l) then addedge(calc(i,j),calc(i+r,j-l));
      if check(i+l,j+r) then addedge(calc(i,j),calc(i+l,j+r));
      if check(i+r,j+l) then addedge(calc(i,j),calc(i+r,j+l));
    end;
end;

procedure work;
var
  i,j:longint;
begin
  for i:=1 to n do
    for j:=1 to m do
      if map[i,j] then begin
        fillchar(chose,sizeof(chose),true);
        if dfs(calc(i,j)) then inc(ans);
      end;
  writeln(n*m-ans);
end;

begin
  into;
  work;
  readln;
  readln;
End.

 

bzoj1191: [HNOI2006]超级英雄Hero

直接跑匈牙利，不能增广就退出就行了。

const
  maxn=1000;

var
  match:array[0..maxn]of longint;
  chose:array[0..maxn]of boolean;
  too:array[0..maxn,1..2]of longint;
  i,j,n,m:longint;

function dfs(x:longint):boolean;
var
  i,j:longint;
begin
  for i:=1 to 2 do begin
    j:=too[x,i];
    if chose[j] then begin
      chose[j]:=false;
      if (match[j]=0) or dfs(match[j]) then begin
        match[j]:=x;
        exit(true);
      end;
    end;
  end;
  exit(false);
end;

begin
  readln(n,m);
  for i:=1 to m do
    readln(too[i,1],too[i,2]);
  for i:=1 to m do begin
    fillchar(chose,sizeof(chose),true);
    if not dfs(i) then begin
      writeln(i-1);
      break;
    end;
    if i=m then writeln(m);
  end;
end.

 

bzoj1854: [Scoi2010]游戏

这题有两个做法，一个是二分图，一个是并查集

二分图，类似超级英雄，直接搞。然后就会tle，问题在于那个fillchar（chose），这个太耗时了，然后就把chose改为记录第几次更新，这样每次就不用fillchar（chose）一边。这是一个新姿势。

type
  arr=record
    toward,next:longint;
  end;

const
  maxm=1000000;
  maxn=20000;

var
  first:array[0..maxn]of longint;
  match,chose:array[0..maxm]of longint;
  edge:array[0..maxm*2]of arr;
  n,tot,sum:longint;



procedure addedge(j,k:longint);
begin
  inc(tot);
  edge[tot].toward:=k;
  edge[tot].next:=first[j];
  first[j]:=tot;
end;

function dfs(x:longint):boolean;
var
  i,too:longint;
begin
  i:=first[x];
  while i>0 do begin
    too:=edge[i].toward;
    if chose[too]<>sum then begin
      chose[too]:=sum;
      if (match[too]=0) or dfs(match[too]) then begin
        match[too]:=x;
        exit(true);
      end;
    end;
    i:=edge[i].next;
  end;
  exit(false);
end;

procedure work;
var
  i,j,k:longint;
begin
  readln(n);
  for i:=1 to n do begin
    readln(j,k);
    addedge(j,i);
    addedge(k,i);
  end;
  fillchar(chose,sizeof(chose),0);
  sum:=1;
  for i:=1 to 10000 do begin
    if not dfs(i) then break;
    inc(sum);
  end;
  writeln(sum-1);
end;


begin
  work;
end.

并查集，具体看hzwer大神，简直吓傻。

http://hzwer.com/2950.html

var
  chose:array[0..10002]of boolean;
  fa:array[0..10002]of longint;
  n,i,j,k,x1,x2:longint;


procedure swap(var x,y:longint);
var
  i:longint;
begin
  i:=x;
  x:=y;
  y:=i;
end;

function find(x:longint):longint;
begin
  if fa[x]<>x then fa[x]:=find(fa[x]);
  exit(fa[x]);
end;

begin
  readln(n);
  for i:=1 to 10002 do fa[i]:=i;
  for i:=1 to n do begin
    readln(j,k);
    x1:=find(j);
    x2:=find(k);
    if x1=x2 then chose[x1]:=true
      else begin
        if x1>x2 then swap(x1,x2);
        fa[x1]:=x2;
        chose[x1]:=true;
      end;
  end;
  for i:=1 to 10001 do
    if not chose[i] then begin
      writeln(i-1);
      break;
    end;
end.

速度上看差不多。

 

bzoj3035: 导弹防御塔

由于保留六位变成保留五位调了一节课才发现我也是醉了。

就是二分时间然后再用匈牙利判断。

type
  arr=record
    toward,next:longint;
  end;
  arr2=record
    x,y:longint;
  end;

const
  maxm=5000000;
  maxn=3000;

var
  edge:array[0..maxm]of arr;
  match:array[0..maxn]of longint;
  first:array[0..100]of longint;
  cost:array[0..maxn]of double;
  chose:array[0..maxn]of boolean;
  map:array[0..100,0..100]of double;
  a,b:array[0..100]of arr2;
  tot,n,m:longint;

procedure addedge(j,k:longint);
begin
  inc(tot);
  edge[tot].toward:=k;
  edge[tot].next:=first[j];
  first[j]:=tot;
end;

function dfs(x:longint):boolean;
var
  i,too:longint;
begin
  i:=first[x];
  while i>0 do begin
    too:=edge[i].toward;
    if not chose[too] then begin
      chose[too]:=true;
      if (match[too]=0) or dfs(match[too]) then begin
        match[too]:=x;
        exit(true);
      end;
    end;
    i:=edge[i].next;
  end;
  exit(false);
end;

function check(x:double):boolean;
var
  i,j,k:longint;
begin
  fillchar(first,sizeof(first),0);
  fillchar(match,sizeof(match),0);
  tot:=0;
  for i:=1 to m do
    for j:=1 to n do
      for k:=1 to n do
        if x-cost[k]-map[i,j]>=0.000000001 then addedge(j,(k-1)*m+i)
          else break;
 // writeln(tot);
  for i:=1 to n do begin
    fillchar(chose,sizeof(chose),false);
    if not dfs(i) then exit(false);
  end;
  exit(true);
end;

procedure into;
var
  i,j,time2,v:longint;
  time1:double;
begin
  readln(m,n,time1,time2,v);
  time1:=time1/60;
  for i:=1 to n do
    read(a[i].x,a[i].y);
  for i:=1 to m do
    read(b[i].x,b[i].y);
  for i:=1 to m do
    for j:=1 to n do
      map[i,j]:=double(sqrt(double(sqr(a[j].x-b[i].x)+sqr(a[j].y-b[i].y)))/v);
 { for i:=1 to m do
    for j:=1 to n do
      writeln(i,' ',j,' ',map[i,j]:0:3);  }
  for i:=1 to n do //begin
    cost[i]:=time1*i+time2*(i-1);
  //  writeln(cost[i]:0:3);
 // end;
end;

procedure work;
var
  left,right,mid:double;
begin
  left:=0;
  right:=maxlongint;
  while right-left>=0.000001 do begin
    mid:=(right+left)/2;
    if check(mid) then right:=mid
      else left:=mid;
  end;
  writeln(left:0:6);
end;

begin
  into;
  work;
end.

 

然后发几道残念！！根本不会写！！我去一定要找时间a掉。

3218: a + b Problem http://www.lydsy.com/JudgeOnline/problem.php?id=3218
2744: [HEOI2012]朋友圈 http://www.lydsy.com/JudgeOnline/problem.php?id=2744
1443: [JSOI2009]游戏Game http://www.lydsy.com/JudgeOnline/problem.php?id=1443
3035: 导弹防御塔http://www.lydsy.com/JudgeOnline/problem.php?id=3035

还有小学生oj若干题，再也不说小学生oj水了，好多题我不会啊啊啊啊！

 

这还怎么参加比赛。