# 程序控

IPPP (Institute of Penniless Peasent-Programmer) Fellow

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Time limit: 3.000 seconds

## 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 1, 2, ..., n as c1, c2, ..., cn where 1 ≤ ci ≤n denotes the ranking of event i in the correct chronological order and a sequence of student responses r1, r2, ..., rn where 1 ≤ ri ≤n 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 2 ≤ n ≤ 20. 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 [1 ... n], 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.

4
4 2 3 1
1 3 2 4
3 2 1 4
2 3 4 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

6
5
10
9

## Analysis

 正确答案 学生答案 含义 输入的序列 3 1 2 4 9 5 10 6 8 7 2 10 1 3 8 4 9 5 7 6 历史事件发生在第几位 转换后的序列 2 3 1 4 6 8 10 9 5 7 3 1 4 6 8 10 9 5 7 2 发生了第几个历史事件

 编号 1 2 3 4 5 6 7 8 9 10 序列 2 3 4 5 6 7 8 9 10 1

 编号 1 2 3 4 5 6 7 8 9 10 序列 5 8 3 7 6 1 9 2 10 4

1. 4在第10位，显然M10 = 1；
2. M9 = 1；
3. 2在第8位，后面只有4比它大，则M8 = M10 + 1 = 2；M7 = M9 + 1 = 2；
4. 1在第6位，后面所有数都比它大，但M7到M10中最大值为M7=M8=2，则M6 = M7(或M8) + 1 = 3；
5. M5 = M7 + 1 = 3；
6. M4 = M7 + 1 = 3；
7. M3 = M4(或M5) + 1 = 4；
8. M2 = M7 + 1 = 3；
9. M1 = M2(或M5) + 1 = 4；

## Solution

#include <iostream>
using namespace std;
int main(void) {
int aTbl, aAns, aMax, nLen, nVar, i = 0;
for (cin >> nLen; i < nLen && cin >> nVar; aTbl[i++] = nVar);
while (true) {
for (i = 0; i < nLen && cin >> nVar; aAns[nVar - 1] = aTbl[i++]);
if (i != nLen) {
break;
}
memset(aMax, 0, sizeof(aMax));
for (nVar = 0, i = nLen - 1; i >= 0; --i) {
for (int j = i + 1; j < nLen; ++j) {
if (aAns[j] > aAns[i] && aMax[j] > aMax[i]) {
aMax[i] = aMax[j];
}
}
nVar = ++aMax[i] > nVar ? aMax[i] : nVar;
}
cout << nVar << endl;
}
return 0;
}
posted on 2010-08-10 15:22  Devymex  阅读(...)  评论(...编辑  收藏