UVA 111(LCS问题)
| 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 point for each event whose rank matches its correct rank
- 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 eventi 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 n integers, indicating the correct chronological order of n events. The remaining lines will each consist of nintegers 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 思路:化简成LCS问题;
正确排序s【】与学生排序p【】的LCS问题:
定义状态转移方程:f【i】【j】为p【】的前i个元素与s【】的前j个元素的LCS长度;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++){
f[i][j]=max(f[i][j-1],f[i-1][j]);
if(s[j]==p[i])
f[i][j]=max(f[i][j],f[i-1][j-1]+1);
}
边界为f[0][0]=0;
注意:输入时s[]与p[]的后缀是事件时间;内容是位置;如果调换了会WA;
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define PI 3.141592653589792128462643383279502
int p[],s[],n;
int f[][];
int main(){
//#ifdef CDZSC_June
freopen("in.txt","r",stdin);
//#endif
//std::ios::sync_with_stdio(false);
cin>>n;
int x;
memset(p,,sizeof(p));
memset(s,,sizeof(s));
for(int i=;i<=n;i++){
cin>>x;s[x]=i;
}
while(!cin.eof()){
for(int i=;i<=n;i++){
cin>>x;p[x]=i;
}
memset(f,,sizeof(f));
for(int i=;i<=n;i++)
for(int j=;j<=n;j++){
f[i][j]=max(f[i-][j],f[i][j-]);
if(s[i]==p[j])
f[i][j]=max(f[i][j],f[i-][j-]+);
}
cout<<f[n][n]<<endl;
}
return ;
}
UVA 111(LCS问题)的更多相关文章
- uva 111 - History Grading (dp, LCS)
题目链接 题意:给N,第二行是答案,n个数c1---cn, 代表第一个的顺序是c1,第二个数顺序是c2; 下面每一行是学生的答案,格式同上. 注意:这个给的顺序需要处理一下,不能直接用. 思路:LCS ...
- UVa 111 History Grading (简单DP,LIS或LCS)
题意:题意就是坑,看不大懂么,结果就做不对,如果看懂了就so easy了,给定n个事件,注意的是, 它给的是第i个事件发生在第多少位,并不是像我们想的,第i位是哪个事件,举个例子吧,4 2 3 1, ...
- uva 111 History Grading(lcs)
题目描述 在信息科学中有一些是关于在某些条件限制下,找出一些计算的最大值. 以历史考试来说好了,学生被要求对一些历史事件根据其发生的年代顺序来排列.所有事件顺序都正确的学生无疑的可以得满分.但是那些没 ...
- uva 111 History Grading(最长公共子序列)
题目连接:111 - History Grading 题目大意:给出一个n 代表序列中元素的个数, 然后是一个答案, 接下来是若干个同学的答案(直到文件结束为止), 求出两个序列的最长公共子序列, 注 ...
- uva 10635 LCS转LIS
这道题两个数组都没有重复的数字,用lcs的nlogn再适合不过了 #include <iostream> #include <string> #include <cstr ...
- UVa 10723 LCS变形 Cyborg Genes
题解转自: UVA 10723 Cyborg Genes - Staginner - 博客园 首先这个题目肯定是按最长公共子序列的形式进行dp的,因为只有保证消去的一部分是最长公共子序列才能保证最后生 ...
- UVA 10534 LCS变种题
求一个序列中 的2*n-1个数字 ,前n+1个数字为严格升序 后n+1个为严格降序,求最长的长度 一开始还没想清楚怎么解,其实就是一个LCS问题,从头到尾以及反序求一下LCS 由于 d[i]为包含了自 ...
- UVA 111 History Grading
读题读了好久,其实就是在输入数据时要对数据的位置进行相应的改变 #include<iostream> #include<cstring> #include<cstdio& ...
- UVa 111 - History Grading (by 最长公共子序列 )
History Grading Background Many problems in Computer Science involve maximizing some measure accor ...
随机推荐
- Sass 基本函数
Sass 中的常用函数 一.字符串函数 1. unquote($string): 删除字符串前后的引号,删除一对引号,如果这个字符串没有带有引号,将返回原始的字符串. 示例: .text1 { con ...
- Jmeter-分布式
转载自: http://www.51testing.com/html/28/116228-247521.html 由于Jmeter本身的瓶颈,当需要模拟数以千计的并发用户时,使用单台机器模拟所有的并发 ...
- Jmeter 自动化脚本录制-Badboy
1. Badboy 录制操作(输入URL其余和平常操作一毛一样). 2. 保存结果,并Export to Jmeter,退出保存. 3. 导入已录制好的文件,添加监听器查看结果. 一般的监听器有(查看 ...
- Java面试基础知识1
1.动态绑定是指在执行期间判断所引用对象的实际类型,根据其实际的类型调用其相应的方法. 2.在将超类转换为子类之前,应该使用instanceof进行检查. 3.包含一个或者多个抽象方法的类本身必须被声 ...
- Python标准库笔记(1) — string模块
String模块包含大量实用常量和类,以及一些过时的遗留功能,并还可用作字符串操作. 1. 常用方法 常用方法 描述 str.capitalize() 把字符串的首字母大写 str.center(wi ...
- vue路由-编程式导航
除了使用 <router-link> 创建 a 标签来定义导航链接,我们还可以借助 router 的实例方法,通过编写代码来实现. router.push(location, onComp ...
- 自己动手实现arm函数栈帧回溯【转】
转自:http://blog.csdn.net/dragon101788/article/details/18668505 内核版本:2.6.14 glibc版本:2.3.6 CPU平台:arm gl ...
- python实战===一键刷屏
#当按键q的时候,自动输入 “大家好!”并回车键发送!from pynput import keyboard from pynput.keyboard import Key, Controller k ...
- 设计模式之笔记--组合模式(Composite)
组合模式(Composite) 定义 组合模式(Composite),将对象组合成树形结构以表示“部分-整体”的层次结构.组合模式使得用户对单个对象和组合对象的使用具有一致性. 组合模式有 ...
- Android SDK更新 Connection to http://dl-ssl.google.com refused
问题: Failed to fetch URL https://dl-ssl.google.com/android/repository/repository-6.xml, reason: Conne ...