HDU - 1160 FatMouse's Speed

 FatMouse's Speed

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

题意：

求出体重递增，速度递减的老鼠，并使其序列长度最长（长度一样的任意输出一组）

重新学习一下回溯路径的方法

代码：

#include
#include
#include
#include
#include
#include
using namespace std;
#define INF 0x3f3f3f3f

struct data
{
	int w,s;
	int num;
}s[1005];

struct data_dp
{
	int sum,pos; //pos用来回溯 
}dp[1005];

bool cmp(data a,data b)
{
	if(a.w == b.w) return a.s >b.s;
	return a.ws[j].w && s[i].s q;//当栈使用 
	q.push_back(k);
	
	while(k != -1)
	{
		q.push_back(dp[k].pos);
		k = dp[k].pos;
	}
	
	// i 的初始化多减的 1为对应的 0 值 
	//倒序输出下标为q[i]的编号 
	for(int i = q.size()-2; i >= 0; i--)
		printf("%d\n",s[q[i]].num);
		
	return 0;
}