Hdoj 1160.FatMouse's Speed 题解

Problem Description

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.

Input

Input contains data for a bunch of mice, one mouse per line, terminated by end of file.

The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.

Two mice may have the same weight, the same speed, or even the same weight and speed.

Output

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 a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that

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

and

S[m[1]] > S[m[2]] > ... > S[m[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 speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one.

Sample Input

6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900

Sample Output

4
4
5
9
7

Source

Zhejiang University Training Contest 2001

---

思路

按照weight↑，speed↓的顺序来做DP，可以先排序使得按照weight↑的顺序，那么问题就转换为求最长递减子序列问题了，详见注释。

代码

#include<bits/stdc++.h>
using namespace std;
struct node
{
	int w;	//表示weight
	int s;	//表示speed
	int id;	//表示序号
}a[1001];
int pre[1001];	//记录前驱位置
int f[1001];
int ans[1001];	//正序存放位置
bool cmp(node x, node y)
{
	if ( x.w == y.w)
		return x.s > y.s;
	return x.w < y.w;
	return false;
}//按照weight↑，speed↓的顺序排
int main()
{
	int x,y;
	int i = 0;
	while(cin>>x>>y)
	{
		a[++i].w = x;
		a[i].s = y;
		a[i].id = i;
		pre[i] = 0;
		f[i] = 1;
	}
	int len = i;
	sort(a+1,a+1+len,cmp);

	int maxlen = -1,maxpos;
	for(int i=1;i<=len;i++)
		for(int j=1;j<i;j++)
		{
			if(a[i].w > a[j].w && a[i].s < a[j].s && f[i] < f[j] + 1)
			{
				f[i] = f[j] + 1;
				pre[i] = j;
				if(f[i] > maxlen)	//求出最大长度顺便记录位置
				{
					maxlen = f[i];
					maxpos = i;
				}
			}
		}
	int pos = maxpos;
	int j = 0;
	while(pos!=0)
	{
		ans[++j] = pos;
		pos = pre[pos];
	}//记录正序的位置顺序
	cout << maxlen << endl;
	for(int i=j;i>=1;i--)
		cout << a[ans[i]].id << endl;
	return 0;
}