codeforces round375(div.2)题解

首先吐槽一下这套题。。。为什么EF全是图论QAQ

过了ABCD四个题。。。F的并查集死磕了好久。。。

不过似乎rank还算乐观。。。（因为ABC都是一次过的QAQ）

Problem A：

啥都不想说QAQ。。。

代码如下：

 var a,b,c,d,max,min,ans:longint;
 begin
   readln(a,b,c);
   max:=a;
   min:=a;
   if (b>max) then max:=b;
   if (c>max) then max:=c;
   if (b<min) then min:=b;
   if (c<min) then min:=c;
   d:=a+b+c-max-min;
   ans:=abs(a-d)+abs(b-d)+abs(c-d);
   writeln(ans);
 end.
   

Problem B：

统计字符串的题，注意一下括号里面的各种细节，以及各种单词统计的处理就可以了。

代码如下：

 var n,i,j,max,ans,now:longint;
     flag:boolean;
     ch:array[..] of char;
 function flagg(x:char):boolean;
 begin
   if (<=ord(x)) and (ord(x)<=+) then exit(true);
   if (<=ord(x)) and (ord(x)<=+) then exit(true);
   exit(false);
 end;
 begin
   readln(n);
   for i:= to n do
     read(ch[i]);
   readln;
   max:=;
   ans:=;
   flag:=true;
   i:=;
   now:=;
   while (i<=n) do
   begin
     if flagg(ch[i]) then inc(now);
     if not(flagg(ch[i])) then
     begin
       if flag then
       begin
         if (now>max) then max:=now;
         now:=;
       end
       else
       begin
         if (now>) then inc(ans);
         now:=;
       end;
     end;
     if (ch[i]='(') then flag:=false;
     if (ch[i]=')') then flag:=true;
     inc(i);
   end;
     if (now>) then
     begin
       if not(flag) then inc(ans);
       if flag and(now>max) then max:=now;
     end;
     writeln(max,' ',ans);
 end.
         

Problem C：

这题的题意需要好好理解一下。。。

首先需要看出，第一个答案就是n div m，

接下来方案只需要按照类似贪心来模拟就可以了，注意一下边界的细节。

代码如下：

 var n,m,i,j,ans,ans1:longint;
     a,flag,flag1,anss:array[..] of longint;
 begin
   readln(n,m);
   fillchar(a,sizeof(a),);
   for i:= to n do
     read(a[i]);
   readln;
   write(n div m,' ');
   ans:=n div m;
   fillchar(flag,sizeof(flag),);
   for i:= to n do
     if (a[i]<=m) then inc(flag[a[i]]);
   ans1:=;
   for i:= to m do
     if (flag[i]<ans) then ans1:=ans1+ans-flag[i];
   writeln(ans1);
   fillchar(flag1,sizeof(flag1),);
   fillchar(anss,sizeof(anss),);
   for i:= to n do
   begin
     if (a[i]<=m) then
     begin
       if (flag1[a[i]]<ans) then
       begin
         inc(flag1[a[i]]);
         anss[i]:=a[i];
       end;
     end;
   end;
   i:=;
   j:=;
   while (j<=m) and (flag1[j]>=ans) do inc(j);
   while (i<=n) do
   begin
     if (anss[i]=) then
     begin
       if (j<>m+) then
       begin
       anss[i]:=j;
       inc(flag1[j]);
       while (j<=m) and (flag1[j]>=ans) do inc(j);
       end else anss[i]:=a[i];
     end;
     inc(i);
   end;
   for i:= to n do
     write(anss[i],' ');
   writeln;
 end.
     

Problem D：

这题题意似乎也没写好QAQ

这题其实就是一个DFS就可以了，由于最开始的内陆湖个数大于等于k，所以直接贪心处较小的那几个内陆湖面积，然后加起来就可以了。

注意细节，比如周围一圈算海洋，不属于内陆湖。

代码如下：

 const tx:array[..] of longint=(,,,-);
     ty:array[..] of longint=(,-,,);
 var n,m,k,i,j,tot,now,ans,ii,jj:longint;
     flag:boolean;
     ch:array[..,..] of char;
     vis:array[..,..] of boolean;
     r:array[..,..] of longint;
     area,b:array[..] of longint;
 procedure qsort(lx,rx:longint);
 var i,j,m,t:longint;
 begin
   i:=lx;
   j:=rx;
   m:=area[(i+j) div ];
   repeat
     while (area[i]<m) do inc(i);
     while (area[j]>m) do dec(j);
     if (i<=j) then
     begin
       t:=area[i];
       area[i]:=area[j];
       area[j]:=t;
       t:=b[i];
       b[i]:=b[j];
       b[j]:=t;
       inc(i);
       dec(j);
     end;
   until (i>j);
   if (i<rx) then qsort(i,rx);
   if (j>lx) then qsort(lx,j);
 end;
 procedure tryit(x,y:longint);
 var i,j,a,b:longint;
 begin
   r[x,y]:=tot+;
   vis[x,y]:=true;
   inc(now);
   if (x=) or (x=n) or (y=) or (y=m) then flag:=false;
   for i:= to  do
   begin
     a:=x+tx[i];
     b:=y+ty[i];
     if (<=a) and (a<=n) and (<=b) and (b<=m) then
       if not(vis[a,b]) and (ch[a,b]='.') then tryit(a,b);
   end;
 end;
 procedure trycolor(t:longint);
 var i,j:longint;
 begin
   for i:= to n do
     for j:= to m do
         if (r[i,j]=t) then ch[i,j]:='*';
 end;
 begin
   readln(n,m,k);
   for i:= to n do
   begin
     for j:= to m do
         read(ch[i,j]);
     readln;
   end;
   fillchar(area,sizeof(area),);
   fillchar(b,sizeof(b),);
   fillchar(r,sizeof(r),);
   tot:=;
   fillchar(vis,sizeof(vis),false);
   for i:= to n- do
     for j:= to m- do
         if (ch[i,j]='.') and not(vis[i,j]) then
         begin
           now:=;
           flag:=true;
           tryit(i,j);
           if flag then
           begin
             inc(tot);
             area[tot]:=now;
           end
           else
           begin
             for ii:= to n do
                 for jj:= to m do
                     if (r[ii,jj]=tot+) then r[ii,jj]:=;
           end;
         end;
   for i:= to tot do
     b[i]:=i;
   qsort(,tot);
   ans:=;
   for i:= to tot-k do
   begin
     trycolor(b[i]);
     ans:=ans+area[i];
   end;
   writeln(ans);
   for i:= to n do
   begin
     for j:= to m do
         write(ch[i,j]);
     writeln;
   end;
 end.

Problem E：

这是一道欧拉混合回路。。。

首先，你需要大胆猜出结论，就是所有度数为偶数的最后都可以满足条件。

然后跑欧拉回路迭代构造方案就可以了。（很传统的做法）

代码如下：

 var t,l,m,n,i,j,ans,u,v,tot:longint;
     deg:array[..] of longint;
     r:array[..,..] of longint;
     vis:array[..] of boolean;
     next,last,other:array[..] of longint;
 procedure insert(x,y:longint);
 begin
   inc(tot);
   next[tot]:=last[x];
   last[x]:=tot;
   other[tot]:=y;
   inc(tot);
   next[tot]:=last[y];
   last[y]:=tot;
   other[tot]:=x;
 end;
 procedure dfs(x:longint);
 var i,j:longint;
 begin
   i:=last[x];
   while (i>) do
   begin
     if not(vis[i div ]) then
     begin
       vis[i div ]:=true;
       dfs(other[i]);
       if (i div <=m) then
       begin
         r[i div ,]:=x;
         r[i div ,]:=other[i];
       end;
     end;
     i:=next[i];
   end;
 end;
 begin
   readln(t);
   for l:= to t do
   begin
     readln(n,m);
     fillchar(vis,sizeof(vis),false);
     fillchar(deg,sizeof(deg),);
     fillchar(r,sizeof(r),false);
     fillchar(next,sizeof(next),);
     fillchar(last,sizeof(last),);
     fillchar(other,sizeof(other),);
     tot:=;
     ans:=;
     for i:= to m do
     begin
       readln(u,v);
       insert(u,v);
       inc(deg[u]);
       inc(deg[v]);
     end;
     for i:= to n do
         if (deg[i] mod =) then insert(i,n+) else inc(ans);
     for i:= to n do
         dfs(i);
     writeln(ans);
     for i:= to m do
         writeln(r[i,],' ',r[i,]);
   end;
 end.
     

Problem F：

本场最细节的题目了。。。

首先，这是一道并查集。。。

然后就是一堆细节了，各种地方需要注意。（比如在第一次联通之后，在清除关系之前需要保留数据）

具体细节都可以看代码。

代码如下：

 var n,m,s,t,ds,dt,i,j,now1,now2,now3,now4,x:longint;
     a,b,father,fatherr,cs,ct:array[..] of longint;
     flag1,flag2,flag,vis:array[..] of boolean;
     flagg,flagt:boolean;
 function tryit(i:longint):longint;
 var k,t,p:longint;
 begin
   k:=i;
   while (father[k]<>k) do k:=father[k];
   t:=i;
   while (t<>k) do
   begin
     p:=father[t];
     father[t]:=k;
     t:=p;
   end;
    exit(k);
 end;
 function tryit1(i:longint):longint;
 var k,t,p:longint;
 begin
   k:=i;
   while (fatherr[k]<>k) do k:=fatherr[k];
   t:=i;
   while (t<>k) do
   begin
     p:=fatherr[t];
     fatherr[t]:=k;
     t:=p;
   end;
    exit(k);
 end;
 procedure mdf(a1,b1:longint);
 var i,j:longint;
 begin
   i:=tryit(a1);
   j:=tryit(b1);
   if (i<>j) then father[j]:=i;
 end;
 begin
   readln(n,m);
   fillchar(a,sizeof(a),);
   fillchar(b,sizeof(b),);
   fillchar(cs,sizeof(cs),);
   fillchar(ct,sizeof(ct),);
   for i:= to m do
     readln(a[i],b[i]);
   readln(s,t,ds,dt);
   for i:= to n do
     father[i]:=i;
   flagg:=false;
   for i:= to m do
   begin
     if (a[i]<>s) and (a[i]<>t) and (b[i]<>s) and (b[i]<>t) then mdf(a[i],b[i]);
     if (a[i]=s) and (b[i]=t) then flagg:=true;
     if (a[i]=t) and (b[i]=s) then flagg:=true;
   end;
   fillchar(flag1,sizeof(flag1),false);
   fillchar(flag2,sizeof(flag2),false);
   fillchar(flag,sizeof(flag),false);
   for i:= to n do
     if (i<>s) and (i<>t) then flag[tryit(i)]:=true;
   for i:= to m do
   begin
     if (a[i]=s) then
     begin
       flag1[tryit(b[i])]:=true;
       cs[tryit(b[i])]:=b[i];
     end;
     if (b[i]=s) then
     begin
       flag1[tryit(a[i])]:=true;
       cs[tryit(a[i])]:=a[i];
     end;
     if (a[i]=t) then
     begin
       flag2[tryit(b[i])]:=true;
       ct[tryit(b[i])]:=b[i];
     end;
     if (b[i]=t) then
     begin
       flag2[tryit(a[i])]:=true;
       ct[tryit(a[i])]:=a[i];
     end;
   end;
   now1:=;
   now2:=;
   now3:=;
   now4:=;
   for i:= to n do
     if (i<>s) and (i<>t) and flag[i] then
     begin
       if flag1[i] and not(flag2[i]) then inc(now1);
       if flag2[i] and not(flag1[i]) then inc(now2);
       if flag1[i] and flag2[i] then inc(now3);
       if not(flag1[i]) and not(flag2[i]) then inc(now4);
     end;
 if (now3>) and ((now1+now2+now3+>ds+dt) or (now1+>ds) or (now2+>dt)) then
 begin
   writeln('No');
   halt;
 end
 else if (now3=) and (not(flagg) or (now1+now2+now3+>ds+dt) or (now1+>ds) or (now2+>dt)) then
 begin
     writeln('No');
     halt;
 end
 else
 begin
   writeln('Yes');
   fatherr:=father;
   fillchar(father,sizeof(father),);
   for i:= to n do
     father[i]:=i;
   flagt:=true;
   for i:= to n do
     if (i<>s) and (i<>t) and flag[i] then
     begin
       if flag1[i] and not(flag2[i]) then
       begin
         dec(ds);
         writeln(s,' ',cs[tryit1(i)]);
         mdf(s,i);
       end
       else if flag2[i] and not(flag1[i]) then
       begin
         dec(dt);
         writeln(t,' ',ct[tryit1(i)]);
         mdf(t,i);
       end;
     end;
   for i:= to n do
     if (i<>s) and (i<>t) and flag[i] then
     begin
       if flag1[i] and flag2[i] then
       begin
         if flagt and (ds>) and (dt>) then
         begin
           dec(ds);
           dec(dt);
           writeln(s,' ',cs[tryit1(i)]);
           writeln(t,' ',ct[tryit1(i)]);
           mdf(s,i);
           mdf(t,i);
           flagt:=false;
         end else
         if (ds>) then
         begin
           dec(ds);
           writeln(s,' ',cs[tryit1(i)]);
           mdf(s,i);
         end else
         begin
           dec(dt);
           writeln(t,' ',ct[tryit1(i)]);
           mdf(t,i);
         end;
       end;
     end;
   if flagg and flagt then
   begin
     dec(ds);
     dec(dt);
     writeln(s,' ',t);
     mdf(s,t);
   end;
   for i:= to m do
     begin
       if (tryit(a[i])<>tryit(b[i])) and (a[i]<>s) and (b[i]<>s) and (a[i]<>t) and (b[i]<>t) then
       begin
         writeln(a[i],' ',b[i]);
         mdf(a[i],b[i]);
       end
     end;
 end;
 end.

完结撒花！～～～