UVA - 111 - History Grading （LCS）

UVA - 111

History Grading

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

Background

Many 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 should partial credit be awarded to students who incorrectly rank one or more of the historical events?

Some possibilities for partial credit include:

1. 1 point for each event whose rank matches its correct rank
2. 1 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 would 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 order of n events  as  where  denotes the ranking of event i in the correct chronological order and a sequence of student responses  where  denotes the chronological rank given by the student to event i; determine the length of 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 integer n indicating the number of events with  . The second line will contain n integers, indicating the correct chronological order of n events. The remaining lines will each consist of n integers with each line representing a student's chronological ordering of the n events. All lines will contain n numbers 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 should print the score for that ranking. There should 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) :: Problem Solving Paradigms :: Dynamic Programming :: Longest Increasing Subsequence (LIS)
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

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思路：题目意思比较难理解......主要解法就是最长公共子序列啦，不过这个题目输入的时候要注意了，直接输入的是每个历史事件发生的时间（先后顺序），相当于给出了每个事件的位置（可以理解为数组的下标），而问你的是填好的历史事件序列中最长公共的序列，所以要先要将每个事件都按时间顺序从小到大排列好（包括正确的答案和学生填的答案），再对事件进行LCS

AC代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>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;}