The Bonus Salary! （poj 3762 离散化+最小费用流）

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The Bonus Salary!

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 2557 Accepted: 675

Description

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_L_i : mm_L_i : ss_L_i to hh_R_i : mm_R_i : ss_R_i.
• 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_L_i:mm_L_i:ss_L_i hh_R_i:mm_R_i:ss_R_i w

Which means, the i-th task must be done from hh_L_i : mm_L_i : ss_L_i to hh_R_i : mm_R_i : ss_R_i and its productivity score is w. (0 ≤hh_L_i, hh_R_i ≤ 23, 0 ≤mm_L_i, mm_R_i, ss_L_i, ss_R_i ≤ 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_R_i : mm_R_i : ss_R_i is strictly later than hh_L_i : mm_L_i : ss_L_i.　

Output

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

Sample Input

5 209:00:00 09:30:00 209:40:00 10:00:00 309:29:00 09:59:00 1009:30:00 23:59:59 407:00:00 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.

Source

题意：有n个任务要在k天时间内选部分完成，每个任务在每天有一个规定的时间，必须要在这个时间内做这个任务，完成一个任务会得到相应的分数，问能得到的最大分数是多少。

思路：离散化+最小费用流，这一题和poj3680是一样的，先对时间离散化，然后添加源点S和汇点T，1->2,2->3,3->4.......T-1->T这样连边，容量为k，费用为0，对于每个任务起始时间u->v连边，容量为1，费用为-w。

代码：

#include <iostream>#include <functional>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i--)#define FRL(i,a,b)  for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")typedef long long ll;using namespace std;#define INF 0x3f3f3f3f#define mod 1000000009const int maxn = 1005;const int MAXN = 4050;const int MAXM = 200010;struct Line{    int a,b,w;}line[MAXN/2];struct Edge{    int to,next,cap,flow,cost;}edge[MAXM];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N,n,k;int a[MAXN],cnt;void init(int n){    N=n;    tol=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int cap,int cost){    edge[tol].to=v;    edge[tol].cap=cap;    edge[tol].cost=cost;    edge[tol].flow=0;    edge[tol].next=head[u];    head[u]=tol++;    edge[tol].to=u;    edge[tol].cap=0;    edge[tol].cost=-cost;    edge[tol].flow=0;    edge[tol].next=head[v];    head[v]=tol++;}bool spfa(int s,int t){    queue<int>q;    for (int i=0;i<N;i++)    {        dis[i]=INF;        vis[i]=false;        pre[i]=-1;    }    dis[s]=0;    vis[s]=true;    q.push(s);    while (!q.empty())    {        int u=q.front();        q.pop();        vis[u]=false;        for (int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if (edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost)            {                dis[v]=dis[u] + edge[i].cost;                pre[v]=i;                if (!vis[v])                {                    vis[v]=true;                    q.push(v);                }            }        }    }    if (pre[t]==-1) return false;    else return true;}//返回的是最大流，cost存的是最小费用int minCostMaxflow(int s,int t,int &cost){    int flow=0;    cost=0;    while (spfa(s,t))    {        int Min=INF;        for (int i=pre[t];i!=-1;i=pre[edge[i^1].to])        {            if (Min > edge[i].cap-edge[i].flow)                Min=edge[i].cap-edge[i].flow;        }        for (int i=pre[t];i!=-1;i=pre[edge[i^1].to])        {            edge[i].flow+=Min;            edge[i^1].flow-=Min;            cost+=edge[i].cost*Min;        }        flow+=Min;    }    return flow;}int main(){#ifndef ONLINE_JUDGE    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);#endif    int i,j,h,m,s;    while (~scanf("%d%d",&n,&k))    {        cnt=0;        for (i=1;i<=n;i++)        {            scanf("%d:%d:%d",&h,&m,&s);            line[i].a=h*3600+m*60+s;            a[++cnt]=line[i].a;            scanf("%d:%d:%d",&h,&m,&s);            line[i].b=h*3600+m*60+s;            a[++cnt]=line[i].b;            scanf("%d",&line[i].w);        }        sort(a+1,a+1+cnt);        cnt=unique(a+1,a+1+cnt)-a-1;        init(cnt+2);        int S=0,T=cnt+1;        for (i=0;i<=cnt;i++)            addedge(i,i+1,k,0);        for (i=1;i<=n;i++)        {            int u=lower_bound(a+1,a+1+cnt,line[i].a)-a;            int v=lower_bound(a+1,a+1+cnt,line[i].b)-a;            addedge(u,v,1,-line[i].w);        }        int ans,cost;        ans=minCostMaxflow(S,T,cost);        printf("%d\n",-cost);    }    return 0;}