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. Mysql 索引原理(转自:张洋)

    摘要 本文以MySQL数据库为 研究对象,讨论与数据库索引相关的一些话题.特别需要说明的是,MySQL支持诸多存储引擎,而各种存储引擎对索引的支持也各不相同,因此MySQL数据 库支持多种索引类型,如 ...

  2. Topcoder SRM 606 div1题解

    打卡! Easy(250pts): 题目大意:一个人心中想了一个数,另一个人进行了n次猜测,每一次第一个人都会告诉他实际的数和猜测的数的差的绝对值是多少,现在告诉你所有的猜测和所有的差,要求你判断心中 ...

  3. C语言.c和.h

    简单的说其实要理解C文件与头文件(即.h)有什么不同之处,首先需要弄明白编译器的工作过程,一般说来编译器会做以下几个过程:       1.预处理阶段 2.词法与语法分析阶段 3.编译阶段,首先编译成 ...

  4. WCF技术剖析 Two

    WCF终结点和寻址之--AddressHead信息匹配代码 Contracts契约 using System; using System.Collections.Generic; using Syst ...

  5. sql 查找表引用的存储过程

    USE [master] GO /****** Object: StoredProcedure [dbo].[uspGetDepends] Script Date: 05/12/2016 14:11: ...

  6. ui_modules和ui_method

    ## 06ui.py #coding:utf-8 import tornado.httpserver import tornado.ioloop import tornado.options impo ...

  7. /proc/sys/shm/drop_caches

    author:skate time:2012/02/22 手工释放linux内存--/proc/sys/vm/drop_cache 转载一篇文章 linux的内存查看: [root@localhost ...

  8. 在C#中调用格式工厂进行任意视频格式到FLV的转换

    1.下载安装格式工厂和完美解码器 2.运行格式工厂,转换一个RMVB到FLV,注意找个大点的,使时间长一些.防止在未执行步骤3时就结束了 3.在进程中查看某个进程的命令行参数是什么? wmic pro ...

  9. HTML+JavaScript制作表白特效,表白不成功,小编现场吃雪

    今年的雪特别美,长沙自从08年后的最大的一场学了,今天小编给大家制作一个表白特效,希望大家喜欢,如果你是程序员希望对你有帮助,追到你喜欢的女孩,哈哈~追不到对象,小编现场吃学给你大家看 下图是爱心飘落 ...

  10. HTML5面向对象的游戏开发简单实例总结

    在阅读一本HTML5游戏开发相关书籍时发现一个很好的例子,通过这个例子可以对面向对象的开发进行更深入的理解.这个对象要实现的是:将一个CSS sprite中的图像绘制到canvas中.首先创建一个Sp ...