题意/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 Sito time Ti. 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 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.
读入/Input：
The 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.
输出/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.
题解/solution：
把一个婚礼分成两个点，分别代表开始时间和结束时间。那么就把问题转换成了一个2-SAT问题。若两个点之间有冲突（也就是时间有重合）则连边。然后就是裸的2-SAT输出问题了。
代码/Code：
const
maxE=2000001;
maxV=3001;
type
arr=record
x,y,next:longint;
end;
var
n,nm,num,tt,tk:longint;
x,y,w,bel,ls,sta,state,dfn,low,du,color,con:array[0..maxV] of longint;
v:array[0..maxV] of boolean;
tu:array[0..maxE] of arr;
function fd(o,p,t,k:longint):boolean;
begin
if (p<=t) or (k<=o) then exit(false);
exit(true);
end;
procedure add(o,p:longint);
begin
inc(nm);
with tu[nm] do
begin
x:=o; y:=p;
next:=ls[o];
ls[o]:=nm;
end;
end;
function min(t,k:longint):longint;
begin
if t<k then exit(t);
exit(k);
end;
procedure init;
var
i,j,h,m:longint;
c1,c2,ct:char;
begin
readln(n);
for i:=1 to n do
begin
read(c1,c2,ct); val(c1+c2,h);
read(c1,c2,ct); val(c1+c2,m);
x[i]:=h*60+m;
read(c1,c2,ct); val(c1+c2,h);
read(c1,c2,ct); val(c1+c2,m);
y[i]:=h*60+m;
readln(w[i]);
end;
for i:=1 to n-1 do
for j:=i+1 to n do
begin
if fd(x[i],x[i]+w[i],x[j],x[j]+w[j]) then
begin
add(i,j+n);
add(j,i+n);
end;
if fd(x[i],x[i]+w[i],y[j]-w[j],y[j]) then
begin
add(i,j);
add(j+n,i+n);
end;
if fd(y[i]-w[i],y[i],x[j],x[j]+w[j]) then
begin
add(i+n,j+n);
add(j,i);
end;
if fd(y[i]-w[i],y[i],y[j]-w[j],y[j]) then
begin
add(i+n,j);
add(j+n,i);
end;
end;
end;
procedure dfs(x:longint);
var
i:longint;
begin
inc(num);
dfn[x]:=num; low[x]:=num;
inc(tt);
sta[tt]:=x; v[x]:=true;
i:=ls[x];
while i>0 do
with tu[i] do
begin
if dfn[y]=0 then
begin
dfs(y);
low[x]:=min(low[x],low[y]);
end else
if v[y] then low[x]:=min(low[x],dfn[y]);
i:=next;
end;
if dfn[x]=low[x] then
begin
inc(tk);
repeat
i:=sta[tt];
dec(tt);
v[i]:=false;
bel[i]:=tk;
until i=x;
end;
end;
procedure tarjan;
var
i:longint;
begin
fillchar(v,sizeof(v),false);
for i:=1 to n*2 do
if dfn[i]=0 then dfs(i);
end;
procedure topsort;
var
head,tail,i:longint;
begin
head:=0; tail:=0;
for i:=1 to tk do
if du[i]=0 then
begin
inc(tail);
state[tail]:=i;
end;
repeat
inc(head);
i:=ls[state[head]];
while i>0 do
with tu[i] do
begin
dec(du[y]);
if du[y]=0 then
begin
inc(tail);
state[tail]:=y;
end;
i:=next;
end;
until head>=tail;
end;
procedure blue(x:longint);
var
i:longint;
begin
color[x]:=-1;
i:=ls[x];
while i>0 do
with tu[i] do
begin
if color[y]=0 then blue(y);
i:=next;
end;
end;
procedure red;
var
i:longint;
begin
for i:=1 to tk do
if color[state[i]]=0 then
begin
color[state[i]]:=1;
blue(con[state[i]]);
end;
end;
procedure print;
var
i,t1,t2:longint;
begin
writeln('YES');
for i:=1 to n do
begin
if color[bel[i]]=1 then
begin
t1:=x[i];
t2:=x[i]+w[i];
end else
begin
t1:=y[i]-w[i];
t2:=y[i];
end;
if t1 div 60<10 then write(0);
write(t1 div 60,':');
if t1 mod 60<10 then write(0);
write(t1 mod 60,' ');
if t2 div 60<10 then write(0);
write(t2 div 60,':');
if t2 mod 60<10 then write(0);
writeln(t2 mod 60);
end;
end;
procedure main;
var
i:longint;
begin
for i:=1 to n do
begin
if bel[i]=bel[i+n] then
begin
writeln('NO');
exit;
end;
con[bel[i]]:=bel[i+n];
con[bel[i+n]]:=bel[i];
end;
fillchar(ls,sizeof(ls),0);
for i:=1 to nm do
with tu[i] do
if bel[x]<>bel[y] then
begin
add(bel[y],bel[x]);
inc(du[bel[x]]);
end;
topsort;
red;
print;
end;
begin
init;
tarjan;
main;
end.