poj 3683(2-sat+拓扑排序）
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 11127 | Accepted: 3785 | 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 Si to 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.
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.
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
Source
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cctype> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<queue> 9 #include<stack> 10 #include<set> 11 #include<vector> 12 #include<algorithm> 13 #include<string.h> 14 typedef long long ll; 15 typedef unsigned long long LL; 16 using namespace std; 17 const int INF=0x3f3f3f3f; 18 const double eps=0.0000000001; 19 const int N=2000+10; 20 const ll mod=1e9+7; 21 struct Node{ 22 int x,y; 23 }a[N]; 24 struct node{ 25 int to,next; 26 }edge[N*N]; 27 struct NODE{ 28 int to,next; 29 }Edge[N*N]; 30 int Head[N]; 31 int head[N],low[N],dfn[N]; 32 int vis[N*10],belong[N*10]; 33 int opp[N*10]; 34 int cnt,t,tot; 35 stack<int>st; 36 void init(){ 37 tot=0; 38 t=0; 39 cnt=0; 40 memset(belong,-1,sizeof(belong)); 41 memset(low,0,sizeof(low)); 42 memset(dfn,0,sizeof(dfn)); 43 memset(vis,0,sizeof(vis)); 44 memset(head,-1,sizeof(head)); 45 } 46 void add(int u,int v){ 47 edge[tot].to=v; 48 edge[tot].next=head[u]; 49 head[u]=tot++; 50 } 51 void ADD(int u,int v){ 52 Edge[tot].to=v; 53 Edge[tot].next=Head[u]; 54 Head[u]=tot++; 55 } 56 void tarjan(int u){ 57 vis[u]=1; 58 st.push(u); 59 dfn[u]=low[u]=++t; 60 for(int i=head[u];i!=-1;i=edge[i].next){ 61 int v=edge[i].to; 62 if(dfn[v]==0){ 63 tarjan(v); 64 low[u]=min(low[u],low[v]); 65 } 66 else if(vis[v]){ 67 low[u]=min(low[u],dfn[v]); 68 } 69 } 70 if(low[u]==dfn[u]){ 71 cnt++; 72 int vv; 73 do{ 74 vv=st.top(); 75 st.pop(); 76 belong[vv]=cnt; 77 vis[vv]=0; 78 }while(vv!=u); 79 } 80 } 81 int n; 82 int in[N]; 83 void topsort(){ 84 queue<int>q; 85 memset(vis,0,sizeof(vis)); 86 for(int i=1;i<=cnt;i++)if(in[i]==0)q.push(i); 87 while(q.empty()==0){ 88 int u=q.front(); 89 q.pop(); 90 if(vis[u]==0){ 91 vis[u]=1; 92 vis[opp[u]]=-1; 93 } 94 for(int i=Head[u];i!=-1;i=Edge[i].next){ 95 int v=Edge[i].to; 96 in[v]--; 97 if(in[v]==0){ 98 q.push(v); 99 } 100 } 101 } 102 } 103 int check(int i,int j){ 104 if(a[i].y<=a[j].x){ 105 return 1; 106 } 107 108 if(a[i].x>=a[j].y){ 109 return 1; 110 } 111 return 0; 112 } 113 char str1[100]; 114 char str2[100]; 115 int main(){ 116 while(scanf("%d",&n)!=EOF){ 117 int val; 118 init(); 119 memset(in,0,sizeof(in)); 120 for(int i=0;i<(n<<1);i=i+2){ 121 scanf("%s%s%d",str1,str2,&val); 122 int x=(str1[0]-'0')*60*10+(str1[1]-'0')*60+(str1[3]-'0')*10+str1[4]-'0'; 123 int y=(str2[0]-'0')*60*10+(str2[1]-'0')*60+(str2[3]-'0')*10+str2[4]-'0';; 124 a[i].x=x; 125 a[i].y=x+val; 126 a[i+1].x=y-val; 127 a[i+1].y=y; 128 } 129 for(int i=0;i<(n<<1);i=i+2){ 130 for(int j=0;j<(n<<1);j=j+2){ 131 if(i==j)continue; 132 if(check(i,j)==0)add(i,j^1); 133 if(check(i,j^1)==0)add(i,j); 134 if(check(i^1,j)==0)add(i^1,j^1); 135 if(check(i^1,j^1)==0)add(i^1,j); 136 } 137 } 138 for(int i=0;i<(n<<1);i++){ 139 if(dfn[i]==0)tarjan(i); 140 } 141 int flag=0; 142 for(int i=0;i<(n<<1);i=i+2){ 143 opp[belong[i]]=belong[i^1]; 144 opp[belong[i^1]]=belong[i]; 145 if(belong[i]==belong[i^1])flag=1; 146 } 147 if(flag){ 148 cout<<"NO"<<endl; 149 continue; 150 } 151 cout<<"YES"<<endl; 152 memset(Head,-1,sizeof(Head)); 153 tot=0; 154 for(int u=0;u<(n<<1);u++){ 155 for(int i=head[u];i!=-1;i=edge[i].next){ 156 int v=edge[i].to; 157 if(belong[u]!=belong[v]){ 158 ADD(belong[v],belong[u]); 159 in[belong[u]]++; 160 } 161 } 162 } 163 164 topsort(); 165 int aa,b,c,d; 166 for(int i=0;i<n;i++){ 167 if(vis[belong[i<<1]]==1){ 168 aa=a[i*2].x/60; 169 b=a[i*2].x%60; 170 c=a[i*2].y/60; 171 d=a[i*2].y%60; 172 //cout<<a[i*2].x<<" "<<a[i*2].y<<endl; 173 printf("%02d:%02d %02d:%02d\n",aa,b,c,d); 174 } 175 else{ 176 aa=a[i*2+1].x/60; 177 b=a[i*2+1].x%60; 178 c=a[i*2+1].y/60; 179 d=a[i*2+1].y%60; 180 //cout<<a[i*2+1].x<<" "<<a[i*2+1].y<<endl; 181 printf("%02d:%02d %02d:%02d\n",aa,b,c,d); 182 } 183 } 184 } 185 }
