ltc男人八题

 【Connected Graph】（ac）

source：pku1737

summarize：这题明显就是dp+推导公式。只考虑第一个点的联通情况。f[i]=pow[i*(i-1) div 2]-  Σ（1，i-1） f[j]*c(i-1,j-1)*pow[(i-j)*(i-j-1) div 2]

program p1737;type row=array[0..100]of longint;var f:Array[0..51]of row; er,d:row; tmp,n:longint; pow:array[0..1226]of row;procedure chen(var a:row;b,c:row);var i,j:longint;begin fillchar(a,sizeof(a),0); for i:=1 to b[0] do for j:=1 to c[0] do inc(a[i+j-1],b[i]*c[j]); a[0]:=i+j-1; for i:=1 to a[0] do begin inc(a[i+1],a[i] div 10000); a[i]:=a[i] mod 10000; end; if a[a[0]+1]>0 then inc(a[0]);end;procedure jian(var a:row;b,c:row);var i:longint;begin fillchar(a,sizeof(a),0); for i:=1 to b[0] do begin a[i]:=b[i]-c[i]; if a[i]<0 then begin a[i]:=a[i]+10000; dec(b[i+1],1); end; end; a[0]:=b[0]; while a[a[0]]=0 do dec(a[0]); if a[0]=0 then inc(a[0]);end;function gcd(x,y:longint):longint;begin if x mod y=0 then exit(y); gcd:=gcd(y,x mod y);end;function c(x,y:longint):row;var i,j:longint; a,b,d:row;begin fillchar(c,sizeof(c),0); c[0]:=1; c[1]:=1; for i:=1 to y do a[i]:=x-i+1; for i:=1 to y do b[i]:=i; for i:=1 to y do if a[i]<>1 then begin for j:=1 to y do if b[j]<>1 then begin tmp:=gcd(a[i],b[j]); a[i]:=a[i] div tmp; b[j]:=b[j] div tmp; if a[i]=1 then break; end; end; fillchar(d,sizeof(d),0); for i:=1 to y do if a[i]<>1 then begin d[0]:=1; d[1]:=a[i]; chen(c,c,d); end;end;procedure predo;var i,j:longint;begin pow[0][0]:=1; pow[0][1]:=1; er[0]:=1; er[1]:=2; for i:=1 to (50*49) div 2 do begin chen(pow[i],pow[i-1],er); end; for i:=1 to 3 do f[i,0]:=1; f[1][1]:=1; f[2][1]:=1; f[3][1]:=4; for i:=4 to 50 do begin f[i]:=pow[i*(i-1) div 2]; for j:=1 to i-1 do begin d:=c(i-1,j-1); chen(d,f[j],d); chen(d,d,pow[(i-j)*(i-j-1) div 2]); jian(f[i],f[i],d); end; end;end;procedure print(n:longint);var i:longint;begin write(f[n,f[n,0]]); for i:=f[n,0]-1 downto 1 do begin if f[n,i]<1000 then write(0); if f[n,i]<100 then write(0); if f[n,i]<10 then write(0); write(f[n,i]); end; writeln;end;begin predo; readln(n); while n<>0 do begin print(n); readln(n); end;end.

 

【Tony's Tour】（wa-ac）

source：pku1739

summarize：无聊的基于联通的状压dp。最后输出的状态时第二位一个前插头，最后一位是尾插头的情况

（一开始wa是因为我把它倒过来做。初始状态是第二位一个前插头，最后一位是一个尾插头，这样是有问题的。）

program Project1;var i,n,m:longint; ch:array[0..100,0..100]of char; pow:Array[0..10]of longint; f,sum:Array[0..1,0..262144]of longint; hash:array[0..262144]of longint;procedure init;var i,j:longint;begin for i:=0 to n+2 do for j:=0 to m+1 do ch[i,j]:='#'; for i:=1 to n do begin for j:=1 to m do read(ch[i,j]); readln; end; for i:=1 to m do ch[n+1,i]:='.';end;function check(X,y:longint):boolean;begin if ch[x,y]='#' then exit(falsE) else exit(true);end;procedure insert(p,x,y:longint);begin if hash[x]>0 then inc(sum[p,hash[x]],y) else begin inc(f[p,0]); sum[p,f[p,0]]:=y; f[p,f[p,0]]:=x; hash[x]:=f[p,0]; end;end;function get(x,k:longint):longint;begin get:=0; if (1 shl (k*2-2))and x>0 then get:=1; if (1 shl (k*2-1))and x>0 then get:=2;end;procedure main;var now,p,l1,i,j,k,x,y,a,b,c,xx:longint; t:boolean;begin p:=0; fillchar(sum,sizeof(sum),0); fillchar(hash,sizeof(hash),0); f[p,1]:=0; f[p,0]:=1; sum[p,1]:=1; for i:=1 to n do begin for j:=1 to m do begin p:=1-p; f[p,0]:=0; for k:=1 to f[1-p,0] do begin x:=f[1-p,k]; y:=sum[1-p,k]; a:=get(x,j); b:=get(x,j+1); if not check(i,j) then insert(p,x,y) else if a+b=0 then begin if check(i,j+1) and check(i+1,j) then begin xx:=x+pow[j-1]+pow[j]*2; insert(p,xx,y); end; end else if (a=1)and(b=2) then begin if (i=n)and(j=m) then begin xx:=x-pow[j-1]*a-pow[j]*b; insert(p,xx,y); end end else if (a>0)and(b=0) then begin if check(i+1,j) then insert(p,x,y); if check(i,j+1) then begin xx:=x-a*pow[j-1]+a*pow[j]; insert(p,xx,y); end end else if (a=0)and(b>0) then begin if check(i,j+1) then insert(p,x,y); if check(i+1,j) then begin xx:=x-b*pow[j]+b*pow[j-1]; insert(p,xx,y); end; end else if (a=2)and(b=1) then begin xx:=x-a*pow[j-1]-b*pow[j]; insert(p,xx,y); end else if (a=1)and(b=1) then begin now:=0; for l1:=j+1 to m do begin c:=get(x,l1); if c=1 then inc(now); if c=2 then dec(now); if now=0 then break; end; xx:=x-pow[j-1]-pow[j]-pow[l1-1]; insert(p,xx,y); end else if (a=2)and(b=2) then begin now:=0; for l1:=j downto 1 do begin c:=get(x,l1); if c=1 then dec(now); if c=2 then inc(now); if now=0 then break; end; xx:=x-pow[j-1]*2-pow[j]*2+pow[l1-1]; insert(p,xx,y); end; end; for k:=1 to f[p,0] do hash[f[p,k]]:=0; end; for j:=1 to f[p,0] do f[p,j]:=f[p,j] shl 2; end; t:=false; for i:=1 to f[p,0] do if f[p,i]=pow[1]+pow[m]*2 then begin t:=true; writeln(sum[p,i]); end; if t=false then writeln(0);end;begin assign(input,'1.in');assign(output,'1.out'); reseT(input);rewrite(output); readln(n,m); pow[0]:=1; for i:=1 to 9 do pow[i]:=pow[i-1] shl 2; while n+m<>0 do begin init; main; readln(n,m); end; close(input);closE(output);end.

 

【A New Stone Game】（ac）

source：pku1740

summarize：这题的方法在我的初学博弈里面有说到，不算难。。

 

program Project1;var n,i,j,tmp:longint; x:array[0..10000]of longint; t:boolean;begin readln(n); while n>0 do begin for i:=1 to n do read(x[i]); for i:=1 to n do for j:=i+1 to n do if x[i]>x[j] then begin tmp:=x[i];x[i]:=x[j];x[j]:=tmp; end; if odd(n) then writeln(1) else begin t:=true; for i:=1 to n div 2 do if x[i*2-1]<>x[i*2] then t:=false; if t then writeln(0) else writeln(1); end; readln(n); end;end.

 

【tree】

source：pku1741

summarize：这题参见了09qzc论文会的。。想不到排序根据单调性来找距离小于m的。

program p1741;const inf=maxlongint shr 2;var root,tot,h,l,st,ans,n,m,i:longint; bj,a,first:array[0..20000]of longint; v:Array[0..20000]of boolean; bian:array[0..100000]of record x,y,c,next:longint;end;procedure connect(a,b,d:longint);begin inc(tot); with bian[tot] do begin X:=a; y:=b; c:=d; nexT:=first[x]; first[x]:=tot; end;end;procedure init;var i,x,y,c:longint;begin readln(n,m); if n=0 then halt; for i:=1 to n do begin first[i]:=0; v[i]:=false; end; tot:=0; st:=0; for i:=1 to n-1 do begin readln(x,y,c); connect(x,y,c); connect(y,x,c); end;end;procedure qsort(l,r:longint);var i,j,tmp,mid:longint;begin mid:=a[(l+r) shr 1]; i:=l;j:=r; repeat while a[i]<mid do inc(i); while a[j]>mid do dec(j); if i<=j then begin tmp:=a[i]; a[i]:=a[j]; a[j]:=tmp; inc(i); dec(j); end; until i>j; if l<j then qsort(l,j); if i<r then qsort(i,r);end;procedure dfs1(x,dep:longint);var k:longint;begin inc(l); a[l]:=dep; k:=first[x]; bj[x]:=st; while k<>0 do with bian[k] do begin if (bj[y]<>st)and (v[y]=false)then dfs1(y,dep+c); k:=next; end;end;function dfs2(x,sum:longint):longint;var tmp,k,min:longint;begin dfs2:=1; tmp:=-1; k:=first[x]; bj[x]:=st; while k<>0 do with bian[k] do begin if (bj[y]<>st)and(v[y]=false)then begin min:=dfs2(y,sum); dfs2:=dfs2+min; if min>tmp then tmp:=min; end; k:=next; end; if (sum-dfs2<=sum div 2)and(tmp<=sum div 2) then begin root:=x; end;end;procedure dfs(x,c:longint);var i,k:longint;begin l:=0; inc(st); dfs1(x,0); if l=1 then exit; if c<>-inf then begin qsort(1,l); h:=1; while (h<l)and(a[h+1]+a[1]+c*2<=m) do inc(h); for i:=1 to l do begin while (h>0)and(a[h]+a[i]+c*2>m) do dec(h); if h<=i then break; dec(Ans,h-i); end; end; inc(st); dfs2(x,l); l:=0; inc(st); dfs1(root,0); qsort(1,l); h:=1; v[root]:=true; while (h<l)and(a[h+1]+a[1]<=m) do inc(h); for i:=1 to l do begin while (h>0)and(a[h]+a[i]>m) do dec(h); if h<=i then break; inc(Ans,h-i); end; k:=first[root]; while k<>0 do with bian[k] do begin if v[y]=false then dfs(y,c); k:=next; end;end;begin while not eof do begin init; ans:=0; dfs(1,-inf); writeln(Ans); end;end.

 【coins】（tle-wa-ac）

source：pku1742

summarize：这题我一开始二进制暴力。tle。。。囧，就写了区间优化的dp。。

program p1742;var ll,rr,lt,rt,a,b,d,c,w:Array[0..100000]of longint; tot,i,l,l2,now,n,m:longint;procedure init;var i:longint;begin readln(n,m); if n+m=0 then halt; for i:=1 to n do read(d[i]); for i:=1 to n do read(c[i]); tot:=0; for i:=1 to n do begin now:=1; while c[i]>now do begin inc(tot); w[tot]:=d[i]*now; c[i]:=c[i]-now; now:=now*2; end; if c[i]>0 then begin inc(tot); w[tot]:=d[i]*c[i]; end; end;end;function max(a,b:longint):longint;begin if a>b then exit(A) else exit(B);end;procedure merge;var i,j,len:longint;begin i:=1;j:=1; a[1]:=0; b[1]:=0; len:=1; while (i<=l)or(j<=l2) do if (i<=l)and((j>l2)or(lt[i]<=ll[j])) then begin if lt[i]<=b[len]+1 then b[len]:=max(b[len],rt[i]) else begin inc(len); a[len]:=lt[i]; b[len]:=rt[i]; end; inc(i); end else begin if ll[j]<=b[len]+1 then b[len]:=max(b[len],rr[j]) else begin inc(len); a[len]:=ll[j]; b[len]:=rr[j]; end; inc(j); end; l:=len; for i:=1 to len do begin lt[i]:=a[i]; rt[i]:=b[i]; end;end;procedure main;var ans,i,k,j:longint;begin l:=1; lt[l]:=0; rt[l]:=0; for k:=1 to tot do begin l2:=0; for i:=1 to l do if lt[i]+w[k]<=m then begin inc(l2); ll[l2]:=lt[i]+w[k]; rr[l2]:=rt[i]+w[k]; if rr[l2]>m then rr[l2]:=m; end; merge; end; ans:=0; for i:=1 to l do ans:=ans+rt[i]-lt[i]+1; dec(ans); writeln(Ans);end;begin while not eof do begin init; main; end;end.

【Musical Theme】

source：pku1743

summarize：这题就是后缀数组+二分答案。。也不难。。

program pku1743;var h,rank,sa,trank,tsa:Array[0..20001]of longint; sum:array[-200..20001]of longint; a,b:Array[0..20001]of longint; n,m:longint;procedure init;var i:longint;begin readln(n); for i:=1 to n do read(b[i]); for i:=1 to n-1 do a[i]:=b[i+1]-b[i]; dec(n); // for i:=1 to n do write(a[i],' ');end;procedure predo;var i,p:longint;begin fillchar(sum,sizeof(sum),0); for i:=1 to n do begin rank[i]:=a[i]; inc(sum[a[i]]); end; for i:=-200 to 200 do sum[i]:=sum[i]+sum[i-1]; for i:=n downto 1 do begin sa[sum[rank[i]]]:=i; deC(sum[rank[i]]); end; trank[sa[1]]:=1; p:=1; for i:=2 to n do begin if rank[sa[i]]<>rank[sa[i-1]] then inc(p); trank[sa[i]]:=p; end;m:=p; rank:=trank;end;procedure premain;var i,j,p:longint;begin j:=1; p:=m; while j<=n do begin fillchar(sum,sizeof(sum),0); p:=0; for i:=n-j+1 to n do begin inc(p); tsa[p]:=i; end; for i:=1 to n do if sa[i]>j then begin inc(p); tsa[p]:=sa[i]-j; end; for i:=1 to n do begin trank[i]:=rank[tsa[i]]; inc(Sum[trank[i]]); end; for i:=2 to m do sum[i]:=sum[i]+sum[i-1]; for i:=n downto 1 do begin sa[sum[trank[i]]]:=tsa[i]; dec(sum[trank[i]]); end; trank[sa[1]]:=1;p:=1; for i:=2 to n do begin if (rank[sa[i]]<>rank[sa[i-1]])or(rank[sa[i]+j]<>rank[sa[i-1]+j]) then inc(p); trank[sa[i]]:=p; end; rank:=trank;m:=p;j:=j*2 end; p:=0; for i:=1 to n do begin if rank[i]=1 then h[1]:=0 else begin j:=sa[rank[i]-1]; while a[i+p]=a[j+p] do inc(p); h[rank[i]]:=p; if p>0 then dec(p); end; end;end;function check(k:longint):boolean;var lmin,rmax,i:longint;begin lmin:=maxlongint;rmax:=0; for i:=2 to n do begin if h[i]<k then begin lmin:=maxlongint;rmax:=0; end else begin if sa[i-1]<lmin then lmin:=sa[i-1]; if sa[i-1]>rmax then rmax:=sa[i-1]; if sa[i]>rmax then rmax:=sa[i]; if sa[i]<lmin then lmin:=sa[i]; end; if rmax-lmin>k then exit(true); end; exit(False);end;procedure work;var ans,mid,l,r:longint;begin l:=0;r:=n div 2; ans:=0; repeat mid:=(l+r)div 2; if check(mid) then l:=mid else r:=mid; until l+1>=r; if check(l) then if l>ans then ans:=l; if check(r) then if r>ans then ans:=r; if ans+1<5 then writeln(0) else writeln(ans+1);end;begin while not eof do begin init; if n=-1 then break; fillchar(h,sizeof(h),0); fillchar(rank,sizeof(rank),0); fillchar(sa,sizeof(sa),0); predo; premain; work; end;end.

【Elevator Stopping Plan】

source：pku1744

summarize：想了很久，很来就每次尽量向上走，二分答案，一遍一遍模拟过去。尽量往上一般会比较优。

program Project1;var a:Array[0..30000]of longint; n:longint; procedure init;var i:longint;begin read(n); if n=0 then halt; for i:=1 to n do read(a[i]);end;function check(x:longint):boolean;var now,st,res,time,bf:longint; t:boolean;begin now:=0; st:=1; time:=0; bf:=1; t:=false; while (now<=n) do begin while (time+abs(a[now+1]-bf)*20<=x)and(now<n) do inc(now); if t then inc(time,10); t:=true; if now=n then exit(true);; st:=a[now+1]; time:=time+(st-bf)*4; if time>x then exit(falsE); res:=0; while time+(st+1-a[now+1])*(20+4)<=x do begin inc(st); inc(res); end; bf:=st; time:=time+res*4; end; if now=n then exit(true) else exit(falsE);end;procedure main;var l,r,mid:longint;begin l:=0;r:=a[n]*20+1; repeat mid:=(l+r) shr 1; if check(mid) then r:=mid else l:=mid; until l+1>=r; writeln(r);end;begin while not eof do begin init; main; end;end.