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UVA - 111 History Grading

时间:2014-11-04 17:28:21      阅读:203      评论:0      收藏:0      [点我收藏+]

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#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[30][30];
int ord[30];
int stu[30];
int main()
{
    //freopen("in","r",stdin);
    int n,i,j,t;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>t;
        ord[t]=i;
    }
    while(cin>>t)
    {
        stu[t]=1;
        for(i=2;i<=n;i++)
        {
            cin>>t;
            stu[t]=i;
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                if(ord[i]==stu[j])
                    dp[i][j]=dp[i-1][j-1]+1;
                else
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
        cout<<dp[n][n]<<endl;
    }
    return 0;
}

UVA - 111
Time Limit:3000MS   Memory Limit:Unknown   64bit IO Format:%lld & %llu

Status

Description

bubuko.com,布布扣


 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 bubuko.com,布布扣 as bubuko.com,布布扣 where bubuko.com,布布扣 denotes the ranking of event i in the correct chronological order and a sequence of student responses bubuko.com,布布扣 where bubuko.com,布布扣 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 bubuko.com,布布扣 . 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 bubuko.com,布布扣 , 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

Status


UVA - 111 History Grading

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原文地址:http://blog.csdn.net/stl112514/article/details/40786815

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