POJ 3762 The Bonus Salary!

The Bonus Salary!

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3762
64-bit integer IO format: %lld      Java class name: Main

 

In order to encourage employees' productivity, ACM Company has made a new policy. At the beginning of a period, they give a list of tasks to each employee. In this list, each task is assigned a "productivity score". After the first K days, the employee who gets the highest score will be awarded bonus salary.

Due to the difficulty of tasks, for task i-th:

• It must be done from hh_Li : mm_Li : ss_Li to hh_Ri : mm_Ri : ss_Ri.
• This range of time is estimated very strictly so that anyone must use all of this time to finish the task.

Moreover, at a moment, each employee can only do at most one task. And as soon as he finishes a task, he can start doing another one immediately.

XYY is very hard-working. Unfortunately, he's never got the award. Thus, he asks you for some optimal strategy. That means, with a given list of tasks, which tasks he should do in the first K days to maximize the total productivity score. Notice that one task can be done at most once.

 

Input

The first line contains 2 integers N and K (1 ≤ N ≤ 2000, 0 ≤ K ≤ 100), indicating the number of tasks and days respectively. This is followed by N lines; each line has the following format:

hh_Li:mm_Li:ss_Li hh_Ri:mm_Ri:ss_Ri w

Which means, the i-th task must be done from hh_Li : mm_Li : ss_Li to hh_Ri : mm_Ri : ss_Ri and its productivity score is w. (0 ≤hh_Li, hh_Ri ≤ 23, 0 ≤mm_Li, mm_Ri, ss_Li, ss_Ri≤ 59, 1 ≤ w ≤ 10000). We use exactly 2 digits (possibly with a leading zero) to represent hh, mm and ss. It is guaranteed that the moment hh_Ri : mm_Ri : ss_Ri is strictly later thanhh_Li : mm_Li : ss_Li.　

 

Output

The output only contains a nonnegative integer --- the maximum total productivity score.

 

Sample Input

5 2
09:00:00 09:30:00 2
09:40:00 10:00:00 3
09:29:00 09:59:00 10
09:30:00 23:59:59 4
07:00:00 09:31:00 3

Sample Output

16

Hint

The optimal strategy is:
Day1: Task1, Task 4
Day2: Task 3
The total productivity score is 2 + 4 + 10 = 16.

解题：最小费用最大流。离散化时间点。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct arc{
     int v,w,f,next;
     arc(int x = ,int y = ,int z = ,int nxt = ){
         v = x;
         w = y;
         f = z;
         next = nxt;
     }
 };
 arc e[];
 int head[maxn],d[maxn],p[maxn],tot,S,T;
 bool in[maxn];
 int n,m,lisan[maxn<<],cnt,x[maxn],y[maxn],sc[maxn];
 void add(int u,int v,int w,int f){
     e[tot] = arc(v,w,f,head[u]);
     head[u] = tot++;
     e[tot] = arc(u,-w,,head[v]);
     head[v] = tot++;
 }
 queue<int>q;
 bool spfa(){
     for(int i = S; i <= T; i++){
         d[i] = INF;
         p[i] = -;
         in[i] = false;
     }
     while(!q.empty()) q.pop();
     d[S] = ;
     in[S] = true;
     q.push(S);
     while(!q.empty()){
         int u = q.front();
         q.pop();
         in[u] = false;
         for(int i = head[u]; ~i; i = e[i].next){
             if(e[i].f >  && d[e[i].v] > d[u] + e[i].w){
                 d[e[i].v] = d[u] + e[i].w;
                 p[e[i].v] = i;
                 if(!in[e[i].v]){
                     in[e[i].v] = true;
                     q.push(e[i].v);
                 }
             }
         }
     }
     return p[T] > -;
 }
 int solve(){
     int tmp = ,minV;
     while(spfa()){
         minV = INF;
         for(int i = p[T]; ~i; i = p[e[i^].v])
             minV = min(minV,e[i].f);
         for(int i = p[T]; ~i; i = p[e[i^].v]){
             tmp += minV*e[i].w;
             e[i].f -= minV;
             e[i^].f += minV;
         }
     }
     return tmp;
 }
 int main(){
     int hh,mm,ss;
     while(~scanf("%d %d",&n,&m)){
         tot = cnt = ;
         memset(head,-,sizeof(head));
         for(int i = ; i < n; i++){
             scanf("%d:%d:%d",&hh,&mm,&ss);
             x[i] = hh* + mm* + ss;
             lisan[cnt++] = x[i];
             scanf("%d:%d:%d",&hh,&mm,&ss);
             y[i] = hh* + mm* + ss;
             lisan[cnt++] = y[i];
             scanf("%d",sc+i);
         }
         sort(lisan,lisan+cnt);
         int cnt1 = ;
         for(int i = ; i < cnt; i++)
         if(lisan[i] != lisan[cnt1-]) lisan[cnt1++] = lisan[i];
         cnt = cnt1;
         S = ;
         T = cnt;
         for(int i = ; i < cnt; i++) add(i,i+,,m);
         for(int i = ; i < n; i++){
             int tx = lower_bound(lisan,lisan+cnt,x[i]) - lisan;
             int ty = lower_bound(lisan,lisan+cnt,y[i]) - lisan;
             add(tx,ty,-sc[i],);
         }
         printf("%d\n",-solve());
     }
     return ;
 }
 /*
 5 2
 09:00:00 09:30:00 2
 09:40:00 10:00:00 3
 09:29:00 09:59:00 10
 09:30:00 23:59:59 4
 07:00:00 09:31:00 3
 */