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

求最长公共子序列

AC代码

#include<stdio.h>
#include<string.h> int max(int a, int b) {
if(a > b)
return a;
else
return b;
} int main() {
int l;
int t;
int num1[100]; scanf("%d", &l);
for(int i = 1; i <= l; i++) {
scanf("%d", &t);
num1[t] = i;
} int num2[100];
while(scanf("%d", &t) != EOF) {
num2[t] = 1;
for(int i = 2; i <= l; i++) {
scanf("%d", &t);
num2[t] = i;
} int dp[100][100];
memset(dp, 0, sizeof(dp)); for(int i = 1; i <= l; i++) {
for(int j = 1; j <= l; j++) {
if(num1[i] == num2[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[l][l]);
} return 0;
}

UVA 111 (复习dp, 14.07.09)的更多相关文章

  1. UVA 674 (入门DP, 14.07.09)

     Coin Change  Suppose there are 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. We ...

  2. UVA 111 简单DP 但是有坑

    题目传送门:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18201 其实是一道不算难的DP,但是搞了好久,才发现原来是题目没 ...

  3. 2019.07.09 纪中_B

    错失AK记 2019.07.09[NOIP提高组]模拟 B 组 明明今天的题都很水,可就是没蒟蒻. 写题的时候: T0一眼高精(结果没切)T1看到2啊8啊果断转二进制观察,发现都是左移几位然后空出的位 ...

  4. LEETCODE 07 09

    最近忙着面试耽误了几天,今天刷了07,09都是字符串处理,一个是大数反转,一个是回文数判断,我都是转成字符串处理的,过了是过了,但是挺慢的,先记着,等有机会优化下 题目 给定一个 32 位有符号整数, ...

  5. UVA.10192 Vacation (DP LCS)

    UVA.10192 Vacation (DP LCS) 题意分析 某人要指定旅游路线,父母分别给出了一系列城市的旅游顺序,求满足父母建议的最大的城市数量是多少. 对于父母的建议分别作为2个子串,对其做 ...

  6. UVA.10130 SuperSale (DP 01背包)

    UVA.10130 SuperSale (DP 01背包) 题意分析 现在有一家人去超市购物.每个人都有所能携带的重量上限.超市中的每个商品有其相应的价值和重量,并且有规定,每人每种商品最多购买一个. ...

  7. 2021.07.09 K-D树

    2021.07.09 K-D树 前置知识 1.二叉搜索树 2.总是很长的替罪羊树 K-D树 建树 K-D树具有二叉搜索树的形态,对于每一个分类标准,小于标准的节点在父节点左边,大于标准的节点在父节点右 ...

  8. 2018.07.09 洛谷P2365 任务安排(线性dp)

    P2365 任务安排 题目描述 N个任务排成一个序列在一台机器上等待完成(顺序不得改变),这N个任务被分成若干批,每批包含相邻的若干任务.从时刻0开始,这些任务被分批加工,第i个任务单独完成所需的时间 ...

  9. 2018.07.09 顺序对齐(线性dp)

    顺序对齐 题目描述 考虑两个字符串右对齐的最佳解法.例如,有一个右对齐方案中字符串是 AADDEFGGHC 和 ADCDEGH. AAD~DEFGGHC ADCDE~~GH~ 每一个数值匹配的位置值 ...

随机推荐

  1. Pandas之DataFrame——Part 3

    ''' [课程2.] 数值计算和统计基础 常用数学.统计方法 ''' # 基本参数:axis.skipna import numpy as np import pandas as pd df = pd ...

  2. windows 下 mySQL 镜像安装文件下载

        前言:有时找到的 MySQL 安装文件是 zip 格式的,需要自己配置,自我感觉麻烦,因此记录下下载镜像安装文件过程. 1. 在浏览器里打开mysql的官网http://www.mysql.c ...

  3. 【ZOJ4061】Magic Multiplication(构造)

    题意:定义一个新运算为两个数A,B上每一位相乘,然后顺次接在一起,现在给定结果C和原来两个数字的长度,要求恢复成原来的数字A,B 若有多解输出A字典序最小的,A相同输出B字典序最小的,无解输出Impo ...

  4. HDU 5159 Card (概率求期望)

    B - Card Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Sta ...

  5. ViewPager实现选项卡功能

    1.ViewPager实现Tab 目录结构:

  6. gdb 打印内存 x

    GNU gdb (Ubuntu -0ubuntu1~ Copyright (C) Free Software Foundation, Inc. License GPLv3+: GNU GPL vers ...

  7. 和菜鸟一起学linux之V4L2摄像头应用流程【转】

    转自:http://blog.csdn.net/eastmoon502136/article/details/8190262/ 上篇文章,知道了,C代码编译后存放在内存中的位置,那么C代码的整个编译过 ...

  8. springBoot【01】

    /* 使用spring官网的 http://start.spring.io/ 来建立项目包 生成入口文件,入口文件中对类注释@SpringBootApplication,这个注释是唯一的,标明这个类是 ...

  9. 配置和读取INI

    #define MAX_FILE_PATH 260 void CControlDlg::OnBnClickedBtnGamepath() { // TODO: 在此添加控件通知处理程序代码 CFile ...

  10. 正则表达式之Regex.Match()用法

    //匹配字符串中的连续数字 string txt = "AAA12345678AAAA"; string m = Regex.Match(txt, @"\d+" ...