UVa 10131 Is Bigger Smarter? (DP&LIS)

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本文介绍了一种解决特定问题的方法，即从一组大象的数据中找出体重递增且智商递减的最大子序列。通过使用C++实现的排序和动态规划算法，解决了这一挑战，并提供了一个示例输入和相应的输出。

10131 - Is Bigger Smarter?

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_problem&problem=1072

Some people think that the bigger an elephant is, the smarter it is. To disprove this, you want to take the data on a collection of elephants and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the IQ's are decreasing.

The input will consist of data for a bunch of elephants, one elephant per line, terminated by the end-of-file. The data for a particular elephant will consist of a pair of integers: the first representing its size in kilograms and the second representing its IQ in hundredths of IQ points. Both integers are between 1 and 10000. The data will contain information for at most 1000 elephants. Two elephants may have the same weight, the same IQ, or even the same weight and IQ.

Say that the numbers on the i-th data line are W[i] and S[i]. Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing an elephant). If these n integers are a[1], a[2],..., a[n] then it must be the case that

   W[a[1]] < W[a[2]] < ... < W[a[n]]

and

   S[a[1]] > S[a[2]] > ... > S[a[n]]

In order for the answer to be correct, n should be as large as possible. All inequalities are strict: weights must be strictly increasing, and IQs must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.

Sample Input

Sample Output

怎么排序？可以按照first和second均从大到小排，first优先。

完整代码：

/*0.016s*/

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1001;

struct node
{
	int wei, iq, id;
	bool operator < (const node& a) const
	{
		return wei > a.wei || wei == a.wei && iq > a.iq;
		///iq也这么排是为了方便后面求dp时不用判断wei是否相等~
	}
} e[MAXN];
int dp[MAXN], path[MAXN];

void print(int x)
{
	printf("%d\n", e[x].id);
	if (path[x]) print(path[x]);
	else printf("%d\n", e[0].id);
}

int main()
{
	int n = 0, ans = 0, i, j;
	while (~scanf("%d%d", &e[n].wei, &e[n].iq))
		e[n].id = n + 1, dp[n++] = 1;
	sort(e, e + n);
	for (i = 1; i < n; ++i)
		for (j = 0; j < i; ++j)
			if (e[j].iq < e[i].iq && dp[j] + 1 > dp[i])
				path[i] = j, dp[i] = dp[j] + 1;
	for (i = 1; i < n; ++i)
		if (dp[ans] < dp[i])
			ans = i;
	printf("%d\n", dp[ans]);
	print(ans);
	return 0;
}