HDOJ 1160 FatMouse's Speed (结构体排序+动态规划)详解

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1460 Accepted Submission(s): 730

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 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

 

Sample Output

44597

 

Source

Zhejiang University Training Contest 2001

 

Recommend

Ignatius

这道题和求解最长单调子序列  Super Jumping! Jumping! Jumping!其实是一样的思路

只不过多了排序。

题意就是：找尽量多的满足   体重越大   速度越慢   的例子。

（居然1A了，做梦一样）

#include<iostream>#include<string>#include<cstring>  #include<cmath>  #include<algorithm>  using namespace std;#define max_n 1010int dp[max_n] = {0};int pos[max_n] = { 0 };//寻找位置  struct mouse{int weight;int speed;int num;//表示位置}mice[max_n];bool cmp(mouse a, mouse b)//重量按照从小到大  速度从大到小{if (a.weight == b.weight)return a.speed > b.speed;return a.weight < b.weight;}int main(){int i = 1;int ans = 0;while (cin>>mice[i-1].weight>>mice[i-1].speed){mice[i-1].num =i;i++;}i--;//因为i多加了1sort(mice, mice + i, cmp);for (int j = 1; j < i; j++)//这个就是按照速度的递减子序列  dp数组里存的是子序列的长度{for (int k = 0; k < i; k++){if (mice[j].weight > mice[k].weight&&mice[j].speed < mice[k].speed)dp[j] = max(dp[j], dp[k] + 1);if (dp[j] > ans)ans = dp[j];}}int p = 0;int t = ans;mouse temp;for (int j = i - 1; j >= 0 && t >= 0; j--)//找到选择的老鼠的位置  t要>=0是为了找到起始位置{if (dp[j] == t&&t == ans) //第一次只要找到最后一个长度最长的老鼠就可以了{t--;pos[p++] = j;temp.weight = mice[j].weight;temp.speed = mice[j].speed;continue;}if (dp[j] == t&&mice[j].weight<temp.weight&&mice[j].speed>temp.speed)//后面的只要满足这个条件就可以选进来{t--;pos[p++] = j;temp.weight = mice[j].weight;temp.speed = mice[j].speed;}}cout << ans +1<< endl;//dp里的长度没有包括起始位置p--;//p多加了1while (p>=0)//因为是倒序存在pos里的{cout << mice[pos[p--]].num << endl;}return 0;}