poj3683(2-SAT 求任意方案)-白红宇

poj3683(2-SAT 求任意方案)

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发布时间：2019-06-20

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基础的2-SAT求任意方案的题目。

Priest John's Busiest Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7309		Accepted: 2492		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

Source

, Dagger and Facer

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            #include 
             
              #include 
              
               #include 
               
                using namespace std;#define INF 0x3fffffff#define N 2020struct node{ int from,to,next;}edge[N*N];int n;int b[N],d[N],k[N];int stk[N],vis[N],low[N],link[N],mark[N];int top,index,id,du[N];//记录入度数int pre[N],cnt,g[N];// g 用来记录排序后的结果int g1[N]; //用来记录缩点后的每一个点所含的点int check(int x,int y,int x1,int y1){ if( y<=x1 || x>=y1 ) return 0; return 1;}void dfs(int s){ mark[s]=1; vis[s]=index++; low[s]=vis[s]; stk[top++]=s; for(int p=pre[s];p!=-1;p=edge[p].next) { int v=edge[p].to; if(mark[v]==0) dfs(v); if(mark[v]==1) low[s]=min(low[s],low[v]); } if(low[s]==vis[s]) { int tmp; id++; do { tmp=stk[top-1]; link[tmp]=id; mark[tmp]=-1; }while(stk[--top]!=s); }}void add_edge(int u,int v){ edge[cnt].from=u; edge[cnt].to=v; edge[cnt].next=pre[u]; pre[u]=cnt++;}void topsort(){ memset(mark,0,sizeof(mark)); top=0; int tcnt=0; for(int i=1;i<=id;i++) if(du[i]==0) { mark[i]=1; stk[top++]=i;//把这个节点入栈 g[tcnt++]=i; } while(top!=0) { int cur=stk[--top]; for(int p=pre[cur];p!=-1;p=edge[p].next) { int v=edge[p].to; if(mark[v]==1) continue; du[v]--; if(du[v]==0) { mark[v]=1; stk[top++]=v; g[tcnt++]=v; } } } //排好了序}void dfs1(int s){ mark[s]=-1; //表示这个点不可以选 for(int p=pre[s];p!=-1;p=edge[p].next) { int v=edge[p].to; dfs1(v); }}int main(){ //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin); //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","w",stdout); while(~scanf("%d",&n)) { cnt=0; memset(pre,-1,sizeof(pre)); for(int i=0;i

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