Ultraviolet-111-History Grading (LCS)

Ultraviolet-111-History Grading (LCS)

Uvs-111 History Grading Time Limit:3000 MS Memory Limit:Unknown 64bit IO Format:% Lld & % llu Submit Status Description Background Using problems in Computer Science involve maximizing some measure according to constraints. Consider a history exam in which students are asked to put several historical events into chronological order. students who order all the events correctly will receive full credit, but how shoshould partial credit be awarded to students who incorrectly rank one or more of the historical events? Some possibilities for partial credit include: 1 point for each event whose rank matches its correct rank1 point for each event in the longest (not necessarily contiguous) sequence of events which are in the correct order relative to each other. For example, if four events are correctly ordered 1 2 3 4 then the order 1 3 2 4 wowould receive a score of 2 using the first method (events 1 and 4 are correctly ranked) and a score of 3 using the second method (event sequences 1 2 4 and 1 3 4 are both in the correct order relative to each other ). In this problem you are asked to write a program to score such questions using the second method. The Problem Given the correct chronological orderNEvents as where denotes the ranking of eventIIn the correct chronological order and a sequence of student responses where denotes the chronological rank given by the student to eventI; DetermineLengthOf the longest (not necessarily contiguous) sequence of events in the student responses that are in the correct chronological order relative to each other. The Input The first line of the input will consist of one integerNIndicating the number of events with. The second line will containNIntegers, indicating the correct chronological orderNEvents. The remaining lines will each consistNIntegers with each line representing a student's chronological ordering of the n events. All lines will containNNumbers in the range, with each number appearing exactly once per line, and with each number separated from other numbers on the same line by one or more spaces. The Output For each student ranking of events your program shocould print the score for that ranking. There shocould be one line of output for each student ranking. Sample Input 1 44 2 3 11 3 2 43 2 1 42 3 4 1 Sample Output 1 123 Sample Input 2 103 1 2 4 9 5 10 6 8 71 2 3 4 5 6 7 8 9 104 7 2 3 10 6 9 1 5 83 1 2 4 9 5 10 6 8 72 10 1 3 8 4 9 5 7 6 Sample Output 2 65109 Source Root: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) Root: aoapc I: Beginning Algorithm Contests (Rujia Liu): Volume 5. Dynamic Programming Root: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim): Problem Solving Paradigms: Dynamic Programming: Longest Increasing Subsequence (LIS) Root: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim): Chapter 3. problem Solving Paradigms: Dynamic Programming: Longest Increasing Subsequence (LIS)-Classical Submit Status

Idea: the meaning of the question is hard to understand ...... the main solution is the longest common subsequence. However, you must note that the input of this question is the time (in sequence) of each historical event ), it is equivalent to giving the location of each event (which can be understood as the subscript of an array), and asking you is the longest common sequence in the historical event sequence, therefore, we must first arrange each event in chronological order from small to large (including correct answers and answers filled by students), and then perform LCS on the event.

AC code:

#include 
 
  #include 
  
   #include 
   
    #include using namespace std;int n, t;int dp[25][25];int a[25];int tmp[25];int main() {cin >> n;for(int i = 0; i < n; i++) {cin >> t;a[t-1] = i;}while(cin >> t) {tmp[t-1] = 0;for(int i = 1; i < n; i++) {cin >> t;tmp[t-1] = i;}memset(dp, 0, sizeof(dp));for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {if(a[j] == tmp[i]) dp[i+1][j+1] = dp[i][j] + 1;else dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j]);}}printf("%d\n", dp[n][n]);}return 0;}