代写CSE 2320、代做C++程序语言、代写dynamic programming、代做c++-代写C/C++编程

CSE 2320 Lab Assignment 2

Due March 5, 2019

Goals:

1. Understanding of the dynamic programming solution to the (one room) weighted interval scheduling problem.

2. Understanding of the five steps for developing a dynamic programming solution.

Requirements:

1. Your task is to write a C program to solve the two room weighted interval scheduling problem by using dynamic

programming. The first line of the input will be the number of intervals (n) and each of the remaining n lines have three

non-negative integers for the start time, finish time, and weight for an interval. The intervals will be ordered

by ascending finish time. You should echo the input.

2. You should print your dynamic programming table(s) with appropriate labeling.

3. Your solution should be formatted as:

a. A line with just the number of intervals assigned to the first room.

b. A line for each of the intervals assigned to the first room: start time, finish time, weight.

c. A line with just the number of intervals assigned to the second room.

d. A line for each of the intervals assigned to the second room: start time, finish time, weight.

e. A line with the total weight that has been achieved

and appear at the end of your output. No particular ordering of the intervals is required.

4. Submit your C program on Blackboard by 3:15 p.m. on March 5 . One of the comment lines should include the

compilation command used on OMEGA (5 point penalty for omitting this).

Getting Started:

1. The input should be read from standard input (which will be one of 1. keyboard typing, 2. a shell redirect (<) from a file,

or 3. cut-and-paste). Do NOT prompt for a file name! Do not use fgets().

2. All output should go to standard output.

3. Your solution must use dynamic programming to maximize the total weight of the intervals assigned to the two rooms.

The intervals assigned to a room may not overlap (two intervals i and j overlap if either

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si < sj < fi or

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si < f j < fi.).

4. Applying the simple (one room) solution in Notes 7.B to get a maximum solution for one room and then applying the same

technique again to the leftover intervals (to assign intervals to the second room) is a non-optimal greedy technique.

5. n will not exceed 50