HDU 1160 FatMouse's Speed 动态规划

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): 6789    Accepted Submission(s): 2960
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

 

解题思路：

用动态规划的思想把有最多元素的满足条件的序列找出来，并且在找的时候把每一种可能的路径保存起来，最后我们就只需输出元素量以及路径就可以了;

把代码中的变量说明一下：

 

首先按照关键字wet升序进行排序再按spd降序排序，

找到满足mou[i].wet > mou[j].wet && mou[i].spd < mou[j].spd条件的，再判断长度，并记录位置

动态规划状态方程为：dp[i] = max(dp[j]+1,dp[i]),1<=j<i;

 

这里定义了一个struct其中的

 

len代表当前元素的状态长度，用来找到找到最长的，

 

id用来记录原始位置。

 

bf用来记录当前元素的前一个元素

解题代码：

// File Name: FatMouse's Speed 1160.cpp
// Author: sheng
// Created Time: 2013年05月14日 星期二 18时28分19秒

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

struct node
{
    int wet, spd;
    int id, len;
    int bf;
    bool operator < (const node a) const
    {
        if (wet == a.wet)
            return spd > a.spd;
        return wet < a.wet;
    }
}tmp, mou[1010];

int main ()
{
    int cun = 1;
    int flag, max = 0;
    int dp[1010];
    int x, y;
    while (scanf ("%d%d", &x, &y) == 2)
    {
        mou[cun] = (node) {x, y, cun, 0, 0};
        cun ++;
    }
    sort(mou+1, mou + cun);
    for (int i = 1; i < cun; i ++)
    {
    //    cout << mou[i].id << "  " << mou[i].wet << "  " << mou[i].spd << endl; 
        for (int j = 1; j < i; j ++)
        {
            if (mou[i].wet > mou[j].wet && mou[i].spd < mou[j].spd)
            {
                if (mou[i].len < mou[j].len + 1)
                {
                    mou[i].len = mou[j].len + 1;
                    mou[i].bf = j;
                    if (mou[i].len >= max)
                    {
                        max = mou[i].len;
                        flag = i;
                    }
                }
            }
        }
    }
    cout << max + 1<< endl;
    dp[1] = mou[flag].id;
    cun = 2;
    while (mou[flag].bf != 0)
    {
        flag = mou[flag].bf;
        dp[cun] = mou[flag].id;
        cun ++;
    }
    for (int i = cun-1; i >= 1; i --)
        cout << dp[i] << endl;
    return 0;
}