UVA-111 History Grading

UVA - 111

History Grading

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

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 eventi; 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

DP求LIS（最长递增子序列）问题。读懂题意很关键:

题目大意：历史成绩评定，给出一些事件的事件先后顺序，然后给出学生的答案，求出得到的先后时间正确的最多个数，以给出成绩。

注意，给出的序列和一般理解上的出现次序是不同的。比如4 2 3 1，一般理解是第4个事件在第1位，第2个事件在第2位，第1个事件在第4位。但在这道题目，序列的含义是第1个事件在第4位，第2个事件在第2位，第4个事件在第一位。为方便计算，需要对输入的序列进行转换，使数字对号入座。比如一个正确的序列是：
     

      
 
正确答案
学生答案
含义
输入的序列
3
 1 2 4 9 5 10 6 8 7
2 10 1 3 8 4 9 5 7 6
历史事件发生在第几位
转换后的序列
2
 3 1 4 6 8 10 9 5 7
3 1 4 6 8 10 9 5 7 2
发生了第几个历史事件
     
接下来就用转换后的序列进行计算。为了方便的看出学生答案的顺序是否正确，应该按照正确答案与1 2 ... 10的对应关系将学生答案再做一次转换。正确答案中第2个历史件事发生在第1位，那么学生答案中的2应转换为1；即3转换为2，以此类推。学生答案就转换为最终的序列为：
     

      
编号
1
 2 3 4 5 6 7 8 9 10
序列
2
 3 4 5 6 7 8 9 10 1
     
代码如下：
#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=25;
int a[maxn],b[maxn],dp[maxn];

int main()
{
    int n,m,i,j,k,s,maxm;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&s);
            a[i]=s;//a[i]=s表示第s年的历史事件是i;
        }
        while(scanf("%d",&s)!=EOF)
        {
            b[s]=a[1];
            maxm=0;
            for(i=2;i<=n;i++)
            {
                scanf("%d",&s);
                b[s]=a[i];
//学生答案（b[]）根据标准答案（a[]）按年份1~n调整相对时间顺序，便于和1~n比较
                     //这里要好好分析，理解
            }            dp[1]=1;            for(i=2;i<=n;i++)//以下是dp关键步骤，从小到大做dp            {                dp[i]=1;                for(j=1;j<i;j++)                if(b[j]<b[i])/*若前面有比b[i]小的，dp[i]取dp[i]和dp[j]+1(包含b[i],所以+1)中较大的*/                dp[i]=max(dp[i],dp[j]+1);                maxm=max(dp[i],maxm);//maxm取dp[i]与之前的maxm中的最大值            }            printf("%d\n",maxm);        }    }    return 0;}

总结：

dp博大精深，代码简洁，想到很难，蕴意深刻。往往是解决问题的利器。