UVA-111 History Grading

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 nintegers, 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

题意：一次历史考试，首先是n，有n个事件，第一行为每个事件发生的正确时间，下面每行是学生提交的事件，对于学生提交的答案，给最多的分数。

        这个题因为提交的每个时间发生的时间，但是最后要找的是学生答对的事件个数，已sample Input2的最后一组数据为例！ 3  1 2 4 9 5 10 6 8 7  这是正确答案，但是它代表的是第一个时间发生在3时刻，第二个事件发生在1时刻，、、、、必须全部转换过来，在寻找学生提交的答案与正确答案的次序。

        要求一个最长公共子序列，如何求最长公共子序列，一个典型的dp问题。

        最长公共子序列的题目博客

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int a[25], b[25], dp[25][25], n, t, i, j;
    while(scanf("%d",&n) != EOF)
    {
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&t);
            a[t] = i;//第个数据在t时刻发生
        }
        while(scanf("%d",&t) != EOF)//判断到结束即可
        {
            b[t] = 1;
            for(i = 2; i <= n; i++)
            {
                scanf("%d",&t);
                b[t] = i;
            }
            // 求两个的最长公共子序列。
            memset(dp, 0, sizeof(dp));
            for(i = 1; i <= n; i++)
            {
                for(j = 1; j <= n; j++)
                {
                    if(a[i] == b[j])
                        dp[i][j] = dp[i - 1][j - 1] + 1;
                    else
                    {
                        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                    }
                }
            }
            printf("%d\n",dp[n][n]);
        }
    }
    return 0;
}