题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1160

FatMouse’s Speed

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 9965 Accepted Submission(s): 4427
Special Judge

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

最长上升子序列及记录路径

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

struct node
{
    int x, y, num;
}p[10010];

bool cmp(node a, node b)
{
    return a.x < b.x;
}

int pre[10010], dp[10010];
vector<int> q;

int main()
{
    int x, y;
    int len = 1;
    while (scanf("%d%d",&x,&y)!=EOF)
    //for (int i = 1; i < 10; i++)
    {
        //cin >> x >> y;
        p[len].x = x;
        p[len].y = y;
        p[len].num = len;
        len++;
    }
    sort(p+1,p+len,cmp);
    memset(pre,0,sizeof(pre));

    for (int i = 1; i < len; i++)
    {
        dp[i] = 1;
        for (int j = 1; j < i; j++)
        {
            if (p[i].x > p[j].x && p[i].y < p[j].y )
            {
                if (dp[i] < dp[j] + 1)
                {
                    dp[i] = dp[j] + 1;
                    pre[p[i].num] = p[j].num;
                }
            }
        }
    }
    int ans = -1, num = 1;
    for (int i = 1; i < len; i++)
    {
        if (ans < dp[i])
        {
            ans = dp[i];
            num = p[i].num;
        }
    }
    printf("%d\n",ans);
    q.clear();
    while (num != 0)
    {
        q.push_back(num);
        num = pre[num];
    }
    reverse(q.begin(),q.end());
    for (int i = 0; i < q.size(); i++)
        printf("%d\n",q[i]);
    return 0;
}