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.

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.

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.

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<string>
  5 using namespace std;
  6 #define MAXN 1020*2
  7 int n;
  8 struct Thing
  9 {
 10     int st,ed,Len;
 11 }p[MAXN];
 12 struct node
 13 {
 14     int v;
 15     node *next;
 16 };
 17 node edge[MAXN*MAXN];
 18 node *cnt=&edge[0];
 19 node *adj[MAXN];
 20 node edge2[MAXN*MAXN];
 21 node *cnt2=&edge2[0];
 22 node *adj2[MAXN];
 23 int dfn[MAXN],low[MAXN],dcnt;
 24 int Belong[MAXN],Num[MAXN];
 25 int stack[MAXN],top;
 26 int In[MAXN],opp[MAXN],scc;
 27 int q[MAXN],col[MAXN];
 28 bool Instack[MAXN],ans[MAXN];
 29 char s1[10],s2[10];
 30 inline void Addedge(int u,int v)
 31 {
 32     node *p=++cnt;
 33     p->v=v;
 34     p->next=adj[u];
 35     adj[u]=p;
 36 }
 37 inline void Addedge2(int u,int v)
 38 {
 39     node *p=++cnt2;
 40     p->v=v;
 41     p->next=adj2[u];
 42     adj2[u]=p;
 43 }
 44 inline int Cal(char str[])
 45 {
 46     int Ret=((str[0]-'0')*10+str[1]-'0')*60+(str[3]-'0')*10+str[4]-'0';
 47     return Ret;
 48 }
 49 inline bool Check(int st1,int len1,int st2,int len2)
 50 {
 51     return (st2<st1+len1)&&(st1<st2+len2);
 52 }
 53 void Read()
 54 {
 55     //scanf("%d",&n);
 56     int i;
 57     for(i=1;i<=n;i++)
 58     {
 59         scanf("%s%s%d",s1,s2,&p[i].Len);
 60         p[i].st=Cal(s1);
 61         p[i].ed=Cal(s2);
 62     }
 63 }
 64 void Build_map()
 65 {
 66     int i,j;
 67     for(i=1;i<=n;i++)
 68         for(j=1;j<=n;j++)
 69         {
 70             if(i==j)
 71                 continue;
 72             if(Check(p[i].st,p[i].Len,p[j].st,p[j].Len))
 73                 Addedge(i,j+n);
 74             if(Check(p[i].st,p[i].Len,p[j].ed-p[j].Len,p[j].Len))
 75                 Addedge(i,j);
 76             if(Check(p[i].ed-p[i].Len,p[i].Len,p[j].st,p[j].Len))
 77                 Addedge(i+n,j+n);
 78             if(Check(p[i].ed-p[i].Len,p[i].Len,p[j].ed-p[j].Len,p[j].Len))
 79                 Addedge(i+n,j);
 80         }
 81 }
 82 void Tarjan(int u)
 83 {
 84     int v;
 85     dfn[u]=low[u]=++dcnt;
 86     stack[++top]=u;
 87     Instack[u]=true;
 88     for(node *p=adj[u];p;p=p->next)
 89     {
 90         v=p->v;
 91         if(!dfn[v])
 92         {
 93             Tarjan(v);
 94             low[u]=min(low[u],low[v]);
 95         }
 96         else if(Instack[v])
 97             low[u]=min(low[u],dfn[v]);
 98     }
 99     if(dfn[u]==low[u])
100     {
101         scc++;
102         do
103         {
104             v=stack[top];
105             top--;
106             Instack[v]=false;
107             Belong[v]=scc;
108             Num[scc]++;
109         }while(v!=u);
110     }
111 }
112 bool Work()
113 {
114     int i;
115     for(i=1;i<=n*2;i++)
116         if(!dfn[i])
117             Tarjan(i);
118     for(i=1;i<=n;i++)
119     {
120         if(Belong[i]==Belong[i+n])
121             return false;
122         opp[Belong[i]]=Belong[i+n];
123         opp[Belong[i+n]]=Belong[i];
124     }
125     int u,v;
126     for(i=1;i<=n*2;i++)
127         for(node *p=adj[i];p;p=p->next)
128         {
129             v=p->v;
130             if(Belong[i]!=Belong[v])
131             {
132                 Addedge2(Belong[v],Belong[i]);
133                 In[Belong[i]]++;
134             }
135         }
136     int l=0,r=0;
137     for(i=1;i<=scc;i++)
138         if(!In[i])
139         {
140             q[r]=i;
141             r++;
142         }
143     while(l<r)
144     {
145         u=q[l];
146         l++;
147         if(!col[u])
148         {
149             col[u]=1;
150             col[opp[u]]=-1;
151         }
152         for(node *p=adj2[u];p;p=p->next)
153         {
154             v=p->v;
155             In[v]--;
156             if(!In[v])
157             {
158                 q[r]=v;
159                 r++;
160             }
161         }
162     }
163     for(i=1;i<=n;i++)
164         if(col[Belong[i]]==1)
165             ans[i]=true;
166     return true;
167 }
168 void Print()
169 {
170     if(Work())
171     {
172         printf("YES\n");
173         int i;
174         for(i=1;i<=n;i++)
175         {
176             if(ans[i])
177                 printf("%02d:%02d %02d:%02d\n",p[i].st/60,p[i].st%60,(p[i].st+p[i].Len)/60,(p[i].st+p[i].Len)%60);
178             else
179                 printf("%02d:%02d %02d:%02d\n",(p[i].ed-p[i].Len)/60,(p[i].ed-p[i].Len)%60,p[i].ed/60,p[i].ed%60);
180         }
181     }
182     else
183         printf("NO\n");
184 }
185 inline void Pre()
186 {
187     memset(edge,0,sizeof(edge));
188     memset(adj,0,sizeof(adj));
189     cnt=&edge[0];
190     memset(edge2,0,sizeof(edge2));
191     memset(adj2,0,sizeof(adj2));
192     cnt2=&edge2[0];
193     memset(dfn,0,sizeof(dfn));
194     memset(low,0,sizeof(low));
195     dcnt=0;
196     top=0;
197     memset(Instack,false,sizeof(Instack));
198     memset(Belong,0,sizeof(Belong));
199     memset(Num,0,sizeof(Num));
200     scc=0;
201     memset(opp,0,sizeof(opp));
202     memset(In,0,sizeof(In));
203     memset(q,0,sizeof(q));
204     memset(col,0,sizeof(col));
205     memset(ans,false,sizeof(ans));
206 }
207 int main()
208 {
209     //freopen("Data.in","r",stdin);
210     //freopen("Data.out","w",stdout);
211     while(scanf("%d",&n)!=EOF)
212     {
213         Pre();
214         Read();
215         Build_map();
216         Print();
217     }
218     return 0;
219 }