The Bonus Salary!
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 ",-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 */