03 2013 档案
摘要：我太弱了。。。[toggle title="太长了只好折叠起来>_a[i]。1934: [Shoi2007]Vote 善意的投票最大流。建立S，T，同意的向S连边，不同意的向T连边，朋友之间连边，流量都是1.跑最小割。2783: [JLOI2012]树树上倍增。枚举每个点i，if (now+sum[a,i]2761: [JLOI2011]不重复数字平衡树。水题，先查找有没有这个数，没有就加入。24...
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摘要：数学期望。总费用=所有边的平均费用和每条边的费用=被优惠且走过的概率*长度，因为长度都是1，所以就是概率。被优惠且走过的概率=优惠路径中包含这条边的概率*走过这条边的概率。=（总包含这条边的路径数/总路径数）^2包含的路径数用两边的点数乘起来就行了。[toggle title="code"][pascal]varn,i,j,k,ee:longint;sum,an,ans:double;now:in...
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摘要：sbt:1503: [NOI2004]郁闷的出纳员 [toggle Title="code "][pascal]varl,r,a,size:Array[1..100000]of longint;le:Array[1..100000]of boolean;ans:int64;root,nn,n,m,i,k,j:longint;c,space:char;procedure lr(var x:longi...
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摘要：这里什么都有，除了code。。。。^_^(其实什么都没有）
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摘要：博客园：http://www.cnblogs.com/lbz007oi/百度空间：http://hi.baidu.com/lbz0072010/cute_loli：http://loli.byethost7.com/
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摘要：求出走到每个点有的方法数a[i]，再求出每个点到终点的方法数b[i]。每条边走的 次数=a[u[i]]*b[v[i]] 。可以用类spfa按照拓扑序求a[i]，把入度为0的点都入栈然后a[e[j]]=a[e[j]]+a[q[h]]求b[i]可以把边反向重新spfa一遍。[toggle Title="code "][pascal]uses math;varu,v,rud,next,e,head:Ar...
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摘要：Welcome to WordPress. This is your first post. Edit or delete it, then start blogging!
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摘要：This is an example page. It's different from a blog post because it will stay in one place and will show up in your site navigation (in most themes). Most people start with an About page that introduc...
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摘要：1 var 2 nnn,root,xx,nn,n,m,i,j:longint;ans:int64; 3 l,r,size,a,an,bn,am,bm:array[1..100000]of longint; 4 function new(x:longint):longint; 5 begin 6 inc(nn); 7 size[nn]:=1;a[nn]:=x; exit(nn); 8 end; 9 10 procedure insert(var t,x:longint); 11 begin 12 if t=0 then 13 begi...
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摘要：1 const inf=10000000; 2 var 3 other,a,dis,pre,uu,vv,ww,next,head,e,x,long,w:array[0..50100]of longint; 4 v:Array[0..10010]of boolean; 5 tt,c,ee,s,t,i,j,ans,mincost,n,m,k:longint; 6 function min(aa,bb:longint):longint; 7 begin 8 if aa>bb then exit(bb) 9 else exit(aa);10 end;11 12 func...
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摘要：1 :loop2 <datamader>.exe3 <slove>.exe4 <program>.exe5 fc <name>.ans <name>.out6 if %errorlevel% == 1 pause7 goto loop程序里没有'<' '>'name是程序名slove是标程program是你的程序注意第5行是.ans 和.out比较datamader不要忘了写 randomize；
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摘要：1 var 2 next,e:array[0..700000]of longint; 3 a,l,r,b:Array[0..150000]of longint; 4 aa,son,d,head,top,point,fa,size,w:array[0..300000]of longint; 5 pp,qq,opt,ll,rr,ansa,ansb,root,uu,vv,i,j,n,ee,xx:longint; 6 v:array[0..300000]of boolean; 7 ch,ch1:char; 8 function max(aa,bb:longin...
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摘要：1 var 2 max,u,v,ans,s,i,j,k,n,m:longint; 3 p,sum:array[0..100000,0..20]of longint; 4 f,d,x:Array[0..100000]of longint; 5 procedure dou(a:longint); 6 var now,i,j,k:longint; 7 begin 8 now:=x[a]; 9 if a=1 then begin if now=s then inc(ans); exit;end;10 k:=trunc(ln(d[a])/ln(2));11 for ...
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摘要：var sum,ee,q,xx,yy,deep,u,c,v,k,i,j,n,m:longint; len,p:array[0..100000,0..20]of longint; a,next,long,e,head,d,f:array[0..100000]of longint;function lca(a,b:longint):longint; var tt,i,j,k:longint; begin if d[a]>d[b] then begin tt:=a;a:=b;b:=tt;end; k:=trunc(ln(d[b])/ln(2)); for i:=k downto 0 ...
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摘要：1 var 2 ans,u,v,x,tot,kk,i,j,k,l,n,m,ee,s,t:longint; 3 d,last,next,long,other,e,head,num:array[0..200000]of longint; 4 5 function min(aa,bb:longint):longint; 6 begin 7 if aa>bb then exit(bb) 8 else exit(aa); 9 end;10 11 procedure add(u,v,c:longint);12 begin13 inc(kk);14 other[kk]...
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