poj3683 pascal题解

Priest John's Busiest Day

Time Limit: 2000MS Memory Limit: 65536KTotal 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

208:00 09:00 3008:15 09:00 20

Sample Output

YES08:00 08:3008:40 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.