poj3683 pascal题解

Priest John's Busiest Day

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

DescriptionJohn 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 toget married on the blessed day. The i-th couple plan to hold their wedding from time Si to time Ti. According to the traditions in the town, there must be a special ceremony on which the couple stand beforethe priest and accept blessings. The i-th couple need Di 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 Si to Si + Di,or from Ti - Di to Ti). 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.InputThe first line contains a integer N ( 1 ≤ N ≤ 1000).The next N lines contain the Si, Ti and Di. Si and Ti are in the format of hh:mm.OutputThe 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

vara: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;beginreadln(m);for i:=1 to m dobeginreadln(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 dojj:=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;beginfor i:=1 to m dobeginif belong[i]>belong[i+m] thenbegina1:=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 elsebegina1:=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;beginf[st[t]]:=false;dec(t);exit(st[t+1]);end;procedure add(x,y:longint);begininc(c[x]);a[x,c[x]]:=y;end;procedure check1;vari:longint;beginfor i:=1 to m doif belong[i]=belong[m+i] thenbeginfff:=false;exit;end;end;function min(x,y:longint):longint;beginif x>y thenexit(y);exit(x);end;procedure check(x,y:longint);beginif not ((ff[x,2]<=ff[y,1]) or (ff[x,1]>=ff[y,2])) thenbeginadd(y+m,x);add(x+m,y);end;if not ((ff[x,4]<=ff[y,3]) or (ff[x,3]>=ff[y,4])) thenbeginadd(y,x+m);add(x,y+m);end;if not ((ff[x,3]>=ff[y,2]) or (ff[x,4]<=ff[y,1])) thenbeginadd(x,y);add(y+m,m+x);end;if not ((ff[x,2]<=ff[y,3]) or (ff[x,1]>=ff[y,4])) thenbeginadd(y,x);add(x+m,y+m);end;end;procedure tarjan(x:longint);vari,y:longint;begininc(d);low[x]:=d;dfn[x]:=d;inc(t);st[t]:=x;f[x]:=true;for i:=1 to c[x] dobeginif not v[a[x,i]] thenbeginv[a[x,i]]:=true;tarjan(a[x,i]);low[x]:=min(low[x],low[a[x,i]]);end elsebeginif f[a[x,i]] thenlow[x]:=min(low[x],dfn[a[x,i]]);end;end;if dfn[x]=low[x] thenbegininc(p);belong[x]:=p;y:=x-1;while x<>y dobeginy:=pp;belong[y]:=p;end;end;end;begininput1;for i:=1 to m dofor j:=i+1 to m docheck(i,j);for i:=1 to 2*m doif not v[i] thenbeginv[i]:=true;tarjan(i);end;fff:=true;check1;if fff thenwriteln('YES') elsebeginwriteln('NO');halt;end;output1;end.