POJ 1513 Scheduling Lectures

Scheduling Lectures

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 1009 Accepted: 433

Description

You are teaching a course and must cover n (1 <= n <= 1000) topics. The length of each lecture is L (1 <= L <= 500) minutes. The topics require t1, t2, ..., tn (1 <= ti <= L) minutes each. For each topic, you must decide in which lecture it should be covered. There are two scheduling restrictions: 
1. Each topic must be covered in a single lecture. It cannot be divided into two lectures. This reduces discontinuity between lectures. 
2. Topic i must be covered before topic i + 1 for all 1 <= i < n. Otherwise, students may not have the prerequisites to understand topic i + 1. 

With the above restrictions, it is sometimes necessary to have free time at the end of a lecture. If the amount of free time is at most 10 minutes, the students will be happy to leave early. However, if the amount of free time is more, they would feel that their tuition fees are wasted. Therefore, we will model the dissatisfaction index (DI) of a lecture by the formula: 

   |0         if t=0         DI=|-C        if 1 <= t <= 10    |(t-10)^2  otherwise      

where C is a positive integer, and t is the amount of free time at the end of a lecture. The total dissatisfaction index is the sum of the DI for each lecture. 

For this problem, you must find the minimum number of lectures that is needed to satisfy the above constraints. If there are multiple lecture schedules with the minimum number of lectures, also minimize the total dissatisfaction index.

Input

The input consists of a number of cases. The first line of each case contains the integer n, or 0 if there are no more cases. The next line contains the integers L and C. These are followed by n integers t1, t2, ..., tn.

Output

For each case, print the case number, the minimum number of lectures used, and the total dissatisfaction index for the corresponding lecture schedule on three separate lines. Output a blank line between cases.

Sample Input

630 1510101010101010120 1080801050302040301201000

Sample Output

Case 1:Minimum number of lectures: 2Total dissatisfaction index: 0Case 2:Minimum number of lectures: 6Total dissatisfaction index: 2700

Source

East Central North America 1998

题目大意:

你在讲述一个课程，它必须涵盖n个主题，每节课的时间为L，每个主题所需的时间分别为t1,t2......tn,对于每个主题，你必须决定它在那一节课内讲授，有两个限制条件:

1.每个主题必须在一节课内授完，不能分成两节课讲授

2.每个主题i必须在主题i+1之间讲授

由于上面的条件，每节课必然有些空闲时间，如果空闲时间不超过10分钟，学生会愉快的离开，如果空闲时间太多，学生会觉得浪费了学费，因此我们用公式为一节课的不满意指数(DI)建模

DI=0(t=0)

DI=-C(1<=t<=10)

DI=(t-10)*(t-10) (t>10)

其中C为一个正整数，t为一节课的空闲时间

你需要找到满足上述限制条件的最少讲课次数，如果有多个解，选择DI值最小的方案

最少讲课次数定义为数组minclass[],最小不满意指数定义为数组minsatisf[]

minclass[i]为主题1~i最少讲课次数，则:

minclass[i]=min(minclass[j])+1(0<=j<i)，本质上就是将j+1和i分到一节课去

于是这段时间的sum:sum+=time[k](i<=k<=j+1),容易发现sum<L

minsatisf[i]=min(minsatisf[j]+DI(L-sum))(0<=j<i)

#include<iostream>#include<cstdio>#include<cstring>using namespace std;inline void _read(int &x){char t=getchar();bool sign=true;while(t<'0'||t>'9'){if(t=='-')sign=false;t=getchar();}for(x=0;t>='0'&&t<='9';t=getchar())x=x*10+t-'0';if(!sign)x=-x;}const int maxn=1005,inf=1e9;int n,L,C,time[maxn],cases;int minclass[maxn],minsatisf[maxn];int DI(int t){if(t==0)return 0;if(t>=1&&t<=10)return -C;if(t>10)return (t-10)*(t-10); }void DP(){int i,j;for(i=1;i<=n;i++){minclass[i]=inf;int sum=0;for(j=i-1;j>=0;j--){sum+=time[j+1];if(sum>L)break;if(minclass[i]<minclass[j]+1)continue;int cost=minsatisf[j]+DI(L-sum);if(minclass[i]==minclass[j]+1&&cost>=minsatisf[i])continue;minclass[i]=minclass[j]+1;minsatisf[i]=cost;}}}void CLEAR(){memset(time,0,sizeof(time));memset(minclass,0,sizeof(minclass));memset(minsatisf,0,sizeof(minsatisf));}int main(){while(scanf("%d",&n)==1&&n){if(cases++)putchar(10);CLEAR();_read(L);_read(C);for(int i=1;i<=n;i++)_read(time[i]);printf("Case %d:\n\n",cases);DP();printf("Minimum number of lectures: %d\n",minclass[n]);printf("Total dissatisfaction index: %d\n",minsatisf[n]);}}