UVA - 111 - History Grading （LCS）

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 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 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

4
4 2 3 1
1 3 2 4
3 2 1 4
2 3 4 1

Sample Output 1

1
2
3

Sample Input 2

10
3 1 2 4 9 5 10 6 8 7
1 2 3 4 5 6 7 8 9 10
4 7 2 3 10 6 9 1 5 8
3 1 2 4 9 5 10 6 8 7
2 10 1 3 8 4 9 5 7 6

Sample Output 2

6
5
10
9

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：題目意思比較難理解......主要解法就是最長公共子序列啦，不過這個題目輸入的時候要注意了，直接輸入的是每個歷史事件發生的時間（先後順序），相當於給出了每個事件的位置（可以理解為數組的下標），而問你的是填好的歷史事件序列中最長公共的序列，所以要先要將每個事件都按時間順序從小到大排列好（包括正確的答案和學生填的答案），再對事件進行LCS

AC代碼：

#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;
}