uva111(dp)

来源:互联网 发布:java transient 作用 编辑:程序博客网 时间:2024/05/20 05:54
A - History Grading
Time Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu
SubmitStatusPracticeUVA 111
Appoint description:

Description

Download as PDF

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 tex2html_wrap_inline34 astex2html_wrap_inline36 wheretex2html_wrap_inline38 denotes the ranking of eventi in the correct chronological order and a sequence of student responsestex2html_wrap_inline42 wheretex2html_wrap_inline44 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 withtex2html_wrap_inline50 . The second line will containn integers, indicating the correct chronological order of n events. The remaining lines will each consist ofn integers with each line representing a student's chronological ordering of the n events. All lines will containn numbers in the range tex2html_wrap_inline60 , 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

44 2 3 11 3 2 43 2 1 42 3 4 1

Sample Output 1

123

Sample Input 2

103 1 2 4 9 5 10 6 8 71 2 3 4 5 6 7 8 9 104 7 2 3 10 6 9 1 5 83 1 2 4 9 5 10 6 8 72 10 1 3 8 4 9 5 7 6

Sample Output 2

65109


看到第二组数据的时候才发现

理解错这道题了

这道题的输入有点坑

4231   这是第一个历史事件发生的顺序是第四  第二个历史时间发生的顺序是第二 ..  第四个历史时间第一个发生

一开始想当然的认为第一个发生的是事件4, 事件2...




#include <stdio.h>#include <string.h>int order[25], arr[25], d[25][25];int max(int x, int y){if (x > y)return x;elsereturn y;}int main(){int n, t;scanf("%d", &n);for (int i = 0; i < n; i++) {scanf("%d", &t);order[t - 1] = i + 1;}while (scanf("%d", &t) != EOF) {arr[t - 1] = 1;for (int i = 1; i < n; i++) {scanf("%d", &t);arr[t - 1] = i + 1; //这才是正确的顺序  输入需要改}memset(d, 0, sizeof(d));for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {if (order[i - 1] == arr[j - 1])d[i][j] = d[i - 1][j - 1] + 1;elsed[i][j] = max(d[i - 1][j], d[i][j - 1]);}printf("%d\n", d[n][n]);}return 0;}
求出最长公共子序列长度 


0 0
原创粉丝点击