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。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 5010;
18 struct arc{
19 int v,w,f,next;
20 arc(int x = 0,int y = 0,int z = 0,int nxt = 0){
21 v = x;
22 w = y;
23 f = z;
24 next = nxt;
25 }
26 };
27 arc e[1000000];
28 int head[maxn],d[maxn],p[maxn],tot,S,T;
29 bool in[maxn];
30 int n,m,lisan[maxn<<4],cnt,x[maxn],y[maxn],sc[maxn];
31 void add(int u,int v,int w,int f){
32 e[tot] = arc(v,w,f,head[u]);
33 head[u] = tot++;
34 e[tot] = arc(u,-w,0,head[v]);
35 head[v] = tot++;
36 }
37 queue<int>q;
38 bool spfa(){
39 for(int i = S; i <= T; i++){
40 d[i] = INF;
41 p[i] = -1;
42 in[i] = false;
43 }
44 while(!q.empty()) q.pop();
45 d[S] = 0;
46 in[S] = true;
47 q.push(S);
48 while(!q.empty()){
49 int u = q.front();
50 q.pop();
51 in[u] = false;
52 for(int i = head[u]; ~i; i = e[i].next){
53 if(e[i].f > 0 && d[e[i].v] > d[u] + e[i].w){
54 d[e[i].v] = d[u] + e[i].w;
55 p[e[i].v] = i;
56 if(!in[e[i].v]){
57 in[e[i].v] = true;
58 q.push(e[i].v);
59 }
60 }
61 }
62 }
63 return p[T] > -1;
64 }
65 int solve(){
66 int tmp = 0,minV;
67 while(spfa()){
68 minV = INF;
69 for(int i = p[T]; ~i; i = p[e[i^1].v])
70 minV = min(minV,e[i].f);
71 for(int i = p[T]; ~i; i = p[e[i^1].v]){
72 tmp += minV*e[i].w;
73 e[i].f -= minV;
74 e[i^1].f += minV;
75 }
76 }
77 return tmp;
78 }
79 int main(){
80 int hh,mm,ss;
81 while(~scanf("%d %d",&n,&m)){
82 tot = cnt = 0;
83 memset(head,-1,sizeof(head));
84 for(int i = 0; i < n; i++){
85 scanf("%d:%d:%d",&hh,&mm,&ss);
86 x[i] = hh*3600 + mm*60 + ss;
87 lisan[cnt++] = x[i];
88 scanf("%d:%d:%d",&hh,&mm,&ss);
89 y[i] = hh*3600 + mm*60 + ss;
90 lisan[cnt++] = y[i];
91 scanf("%d",sc+i);
92 }
93 sort(lisan,lisan+cnt);
94 int cnt1 = 1;
95 for(int i = 1; i < cnt; i++)
96 if(lisan[i] != lisan[cnt1-1]) lisan[cnt1++] = lisan[i];
97 cnt = cnt1;
98 S = 0;
99 T = cnt;
100 for(int i = 0; i < cnt; i++) add(i,i+1,0,m);
101 for(int i = 0; i < n; i++){
102 int tx = lower_bound(lisan,lisan+cnt,x[i]) - lisan;
103 int ty = lower_bound(lisan,lisan+cnt,y[i]) - lisan;
104 add(tx,ty,-sc[i],1);
105 }
106 printf("%d\n",-solve());
107 }
108 return 0;
109 }
110 /*
111 5 2
112 09:00:00 09:30:00 2
113 09:40:00 10:00:00 3
114 09:29:00 09:59:00 10
115 09:30:00 23:59:59 4
116 07:00:00 09:31:00 3
117 */