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 = 5010;
struct arc{
    int v,w,f,next;
    arc(int x = 0,int y = 0,int z = 0,int nxt = 0){
        v = x;
        w = y;
        f = z;
        next = nxt;
    }
};
arc e[1000000];
int head[maxn],d[maxn],p[maxn],tot,S,T;
bool in[maxn];
int n,m,lisan[maxn<<4],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,0,head[v]);
    head[v] = tot++;
}
queue<int>q;
bool spfa(){
    for(int i = S; i <= T; i++){
        d[i] = INF;
        p[i] = -1;
        in[i] = false;
    }
    while(!q.empty()) q.pop();
    d[S] = 0;
    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 > 0 && 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] > -1;
}
int solve(){
    int tmp = 0,minV;
    while(spfa()){
        minV = INF;
        for(int i = p[T]; ~i; i = p[e[i^1].v])
            minV = min(minV,e[i].f);
        for(int i = p[T]; ~i; i = p[e[i^1].v]){
            tmp += minV*e[i].w;
            e[i].f -= minV;
            e[i^1].f += minV;
        }
    }
    return tmp;
}
int main(){
    int hh,mm,ss;
    while(~scanf("%d %d",&n,&m)){
        tot = cnt = 0;
        memset(head,-1,sizeof(head));
        for(int i = 0; i < n; i++){
            scanf("%d:%d:%d",&hh,&mm,&ss);
            x[i] = hh*3600 + mm*60 + ss;
            lisan[cnt++] = x[i];
            scanf("%d:%d:%d",&hh,&mm,&ss);
            y[i] = hh*3600 + mm*60 + ss;
            lisan[cnt++] = y[i];
            scanf("%d",sc+i);
        }
        sort(lisan,lisan+cnt);
        int cnt1 = 1;
        for(int i = 1; i < cnt; i++)
        if(lisan[i] != lisan[cnt1-1]) lisan[cnt1++] = lisan[i];
        cnt = cnt1;
        S = 0;
        T = cnt;
        for(int i = 0; i < cnt; i++) add(i,i+1,0,m);
        for(int i = 0; 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],1);
        }
        printf("%d
",-solve());
    }
    return 0;
}
/*
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
*/