Clover 杯NOIP模拟赛 II

Freda的烦恼

　　 CF原题http://codeforces.com/contest/215/problem/D

　　贪心，只租一辆或者让所有人不要求赔偿。花费关于满载车辆是一个一次函数，题目相当于一次函数在一个区间里求最大值。

Code
type Pointer1=^node1;
     node1=record
       v:integer;
       data:char;
       next:Pointer1;
       end;
     Pointer2=^node2;
     node2=record
       v:integer;
       next:Pointer2;
       end;
var e:array[1..100] of Pointer1;
    map:array[1..100,'A'..'z'] of Pointer2;
    Q,s,t,n,m,i,ans:longint;
    c,ttt:char;
    a:array[1..100] of integer;
procedure build(s,t:longint;c:char);
var P:Pointer1;
begin new(p);p^.v:=t;p^.data:=c;p^.next:=e[s];e[s]:=p;end;
procedure add(s:longint;c:char;t:longint);
var q:Pointer2;
begin new(q);q^.v:=t;q^.next:=map[s][c];map[s][c]:=q;end;
procedure BFS(x,y:integer):boolean;
          var
          begin
          h:=0;t:=1;Que[1]:=x;

          end;
begin
assign(input,'trav.in');
assign(output,'trav.out');
reset(input);rewrite(output);
readln(N,M);ans:=maxlongint;
for i:=1 to M do
  begin
  readln(s,t,ttt,c);
  build(s,t,c);
  add(t,c,s);
  end;
readln(Q);
for i:=1 to q do read(a[i]);
for i:=1 to q-1 do
  begin
  ans:=maxlongint;
  BFS(a[i],a[i+1],0);
  if ans<>maxlongint then writeln(ans)
                     else writeln('-1');
  end;
close(input);close(output);
end.

Freda的迷宫

　　 std是求桥，有人说LCA暴力可过。明天学一下这俩东西。

　　学过Tarjan发明的算法的都说Tarjan神。。。一次DFS遍历能求桥求割点求LCA，代码差不多，连并查集都跟他有关系，太厉害了。

两点间有唯一路径等价于两点间在原图中任意路径上不存在非桥边，所以我们把桥找到，然后在新的图中只保留桥，利用并查集判断两点间连通性。这里具体实现利用边是否访问过来保证拓扑。

Code
uses math;
const Max_N=10000;
      Max_M=100000;
var v,next:array[1..Max_M*2] of longint;
    DFN,LOW,Adjlist,father:array[1..Max_N] of longint;
    InS:array[1..Max_N] of boolean;
    bridge,vis:array[1..Max_M] of boolean;
    l,r:array[1..Max_M] of longint;
    tot,i,m,n,q,Index,x,y:longint;
procedure build(x,y:integer);
          begin
          inc(tot);v[tot]:=y;
          next[tot]:=Adjlist[x];
          Adjlist[x]:=tot;
          end;
function  find(x:longint):longint;
          begin
          if father[x]=x then exit(x);
          father[x]:=find(father[x]);
          find:=father[x];
          end;
procedure union(x,y:longint);
begin father[find(x)]:=find(y);end;
procedure Tarjan(u:longint);
          var i:longint;
          begin
          inc(Index);
          DFN[u]:=Index;
          LOW[u]:=Index;
          InS[u]:=true;
          i:=Adjlist[u];
          while i<>0 do
            begin
            if not vis[i>>1] then
              begin
              vis[i>>1]:=true;
              if DFN[v[i]]=0
                then begin
                     Tarjan(v[i]);
                     LOW[u]:=min(LOW[u],LOW[v[i]]);
                     if DFN[u]<LOW[v[i]]
                       then bridge[i>>1]:=true;
                     end
                else begin
                     if InS[v[i]] then
                        LOW[u]:=min(DFN[v[i]],LOW[u]);
                     end;
              end;
            i:=next[i];
            end;
          InS[u]:=false;
          end;
begin
readln(n,m,q);tot:=1;
for i:=1 to n do
  father[i]:=i;
for i:=1 to m do
  begin
  readln(l[i],r[i]);
  build(l[i],r[i]);
  build(r[i],l[i]);
  end;
for i:=1 to n do
  if DFN[i]=0
    then Tarjan(i);
for i:=1 to N do
  if bridge[i]
    then union(l[i],r[i]);
for i:=1 to Q do
  begin
  readln(x,y);
  if find(x)=find(y)
    then writeln('Y')
    else writeln('N');
  end;
end.

Freda的旗帜

Code
var s:string;
    j,k,count:longint;
    A:array[1..10,1..10] of boolean;
    ans,i,g,n,c:Qword;
    d:array[1..10] of longint;
    f1,f2:array[1..10] of qword;
    f:boolean;
function UPint(x:real):qword;
         begin
         UPint:=trunc(x);
         if x>trunc(x) then inc(UPint);
         end;
begin
readln(n,c);
for j:=1 to c do
  begin
  readln(s);
  for k:=1 to c do
    if s[k]='1'
      then begin
           A[j][k]:=true;
           inc(d[j]);
           end;
  end;
f:=true;
for j:=1 to c do
  begin
  if d[j]>1 then f:=false;
  if d[j]>0 then inc(count);
  end;
if f then begin
          dec(n,c);
          ans:=1;
          inc(ans,UPint(n/count));
          writeln(ans);
          halt;
          end;
for j:=1 to c do f1[j]:=1;
g:=c;i:=2;
while i<=n do
  begin
  for j:=1 to c do
    begin
    for k:=1 to c do
      if A[j][k]then
        inc(f2[j],f1[k]);
    inc(G,f2[j]);
    if G>=n
      then begin
           writeln(i);
           halt;
           end;
    end;
  f1:=f2;fillchar(f2,sizeof(f2),0);
  i:=i+1;
  end;
end.

总结

110分75名，这次名次好低啊。我应该写第二题暴力，唉。还是应该注意不能AC就写部分数据。毕竟noip分数线去年不到400，也就是说一次模拟赛A掉第一题，二三题加起来能的一百分就行了。学会骗分。