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。
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。
时间： 2024-10-12 14:09:23