poj3683 pascal题解

Priest John's Busiest Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8237		Accepted: 2801		Special Judge

Description

John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time S_i to time T_i. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. The i-th couple need D_i minutes to finish this ceremony. Moreover, this ceremony must be either at the beginning or the ending of the wedding (i.e. it must be either from S_i to S_i + D_i, or from T_i - D_i to T_i). Could you tell John how to arrange his schedule so that he can present at every special ceremonies of the weddings.

Note that John can not be present at two weddings simultaneously.

Input

The first line contains a integer N ( 1 ≤ N ≤ 1000). 
The next N lines contain the S_i, T_i and D_i. S_i and T_i are in the format of hh:mm.

Output

The first line of output contains "YES" or "NO" indicating whether John can be present at every special ceremony. If it is "YES", output another N lines describing the staring time and finishing time of all the ceremonies.

Sample Input

2
08:00 09:00 30
08:15 09:00 20

Sample Output

YES
08:00 08:30
08:40 09:00

var
  a:array [1..3010,0..3010] of longint;
  v,f:array [1..3010] of boolean;
  low,dfn,st,belong,c:Array [1..3010] of longint;
  ff:array [1..3010,1..4] of longint;
  a1,a2,a3,a4,i,j,m,n,x,y,mm,nn,p,ii,s2,len,s1,d,jj,t,k:longint;
  ch:char;
  fff:boolean;
  s:string;


procedure input1;
begin
  readln(m);
  for i:=1 to m do
  begin
    readln(s);
    len:=length(s);
    jj:=0;
    x:=ord(s[1])*10+ord(s[2])-ord('0')*11;
    y:=ord(s[4])*10+ord(s[5])-ord('0')*11;
    mm:=ord(s[7])*10+ord(s[8])-ord('0')*11;
    nn:=ord(s[10])*10+ord(s[11])-ord('0')*11;
    for j:=13 to len do
    jj:=jj*10+ord(s[j])-ord('0');
    s1:=x*60+y;
    s2:=mm*60+nn;
    ff[i,1]:=s1;
    ff[i,2]:=s1+jj;
    ff[i,3]:=s2-jj;
    ff[i,4]:=s2;
  end;
end;


procedure output1;
begin
  for i:=1 to m do
  begin
    if belong[i]>belong[i+m] then
    begin
      a1:=ff[i,1] div 60;
      a2:=ff[i,1] mod 60;
      a3:=ff[i,2] div 60;
      a4:=ff[i,2] mod 60;
      writeln(a1 div 10,a1 mod 10,':',a2 div 10,a2 mod 10,' ',a3 div 10,a3 mod 10,':',a4 div 10,a4 mod 10);
    end else
    begin
      a1:=ff[i,3] div 60;
      a2:=ff[i,3] mod 60;
      a3:=ff[i,4] div 60;
      a4:=ff[i,4] mod 60;
      writeln(a1 div 10,a1 mod 10,':',a2 div 10,a2 mod 10,' ',a3 div 10,a3 mod 10,':',a4 div 10,a4 mod 10);
    end;
  end;
end;


function pp:longint;
begin
  f[st[t]]:=false;
  dec(t);
  exit(st[t+1]);
end;


procedure add(x,y:longint);
begin
  inc(c[x]);
  a[x,c[x]]:=y;
end;


procedure check1;
var
  i:longint;
begin
  for i:=1 to m do
  if belong[i]=belong[m+i] then
  begin
    fff:=false;
    exit;
  end;
end;


function min(x,y:longint):longint;
begin
  if x>y then
  exit(y);
  exit(x);
end;


procedure check(x,y:longint);
begin
  if not ((ff[x,2]<=ff[y,1]) or (ff[x,1]>=ff[y,2])) then
  begin
    add(y+m,x);
    add(x+m,y);
  end;
  if not ((ff[x,4]<=ff[y,3]) or (ff[x,3]>=ff[y,4])) then
  begin
    add(y,x+m);
    add(x,y+m);
  end;
  if not ((ff[x,3]>=ff[y,2]) or (ff[x,4]<=ff[y,1])) then
  begin
    add(x,y);
    add(y+m,m+x);
  end;
  if not ((ff[x,2]<=ff[y,3]) or (ff[x,1]>=ff[y,4])) then
  begin
    add(y,x);
    add(x+m,y+m);
  end;
end;


procedure tarjan(x:longint);
var
  i,y:longint;
begin
  inc(d);
  low[x]:=d;
  dfn[x]:=d;
  inc(t);
  st[t]:=x;
  f[x]:=true;
  for i:=1 to c[x] do
  begin
    if not v[a[x,i]] then
    begin
      v[a[x,i]]:=true;
      tarjan(a[x,i]);
      low[x]:=min
      (low[x],low[a[x,i]]);
    end else
    begin
      if f[a[x,i]] then
      low[x]:=min(low[x],dfn[a[x,i]]);
    end;
  end;
  if dfn[x]=low[x] then
  begin
    inc(p);
    belong[x]:=p;
    y:=x-1;
    while x<>y do
    begin
      y:=pp;
      belong[y]:=p;
    end;
  end;
end;




begin
  input1;
  for i:=1 to m do
    for j:=i+1 to m do
    check(i,j);
  for i:=1 to 2*m do
  if not v[i] then
  begin
    v[i]:=true;
    tarjan(i);
  end;
  fff:=true;
  check1;
  if fff then
  writeln('YES') else
  begin
    writeln('NO');
    halt;
  end;
  output1;
end.