FatMouse‘s Speed

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

一组数据，处理到文件结束

给你一些牛的体重，和速度

现在从这些牛中挑出尽量多的牛组成一个序列，使得满足：

体重从小到大严格递减，速度从大到小严格递增

输出最多能找出多少只牛，还要输出牛的序号

其实类似最大上升子序列一样，只不过有2个条件，一个增大，一个减少而已。

再另外记录下原来输入的牛的编号，

这样排序后照样可以知道他原来的序号。

按原来的序号输出

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cstdio>
 5
 6 using namespace std;
 7
 8 const int maxn=1005;
 9
10 int dp[maxn];                    //以第i个为结束元素的序列的最大值
11 int pre[maxn];                   //记录序列
12
13 struct Node
14 {
15     int w,s,num;                 //num记录输入的顺序
16 }node[maxn];
17
18 bool cmp(Node a,Node b)
19 {
20     if(a.w==b.w)
21         return a.s>b.s;
22     return a.w<b.w;
23 }
24
25 //输出路径
26 void output(int cur)
27 {
28     if(pre[cur]!=-1)
29         output(pre[cur]);
30     cout<<node[cur].num<<endl;
31 }
32
33 int main()
34 {
35     //freopen("in.txt","r",stdin);          //记住要注释掉
36
37     int u,v;
38     int tot=1;
39
40     while(cin>>u>>v)
41     {
42         node[tot].w=u;
43         node[tot].s=v;
44         node[tot].num=tot++;
45     }
46
47     sort(node+1,node+tot,cmp);
48
49     memset(pre,-1,sizeof(pre));
50
51     for(int i=1;i<tot;i++)
52         dp[i]=1;
53
54     for(int i=1;i<tot;i++)
55     {
56         for(int j=1;j<i;j++)
57             if(node[j].w<node[i].w&&node[j].s>node[i].s)
58                 if(dp[j]+1>dp[i])
59                 {
60                     dp[i]=dp[j]+1;
61                     pre[i]=j;
62                 }
63     }
64
65     int ans=0;
66     int cur;
67     for(int i=1;i<tot;i++)
68         if(dp[i]>ans)
69         {
70             ans=dp[i];
71             cur=i;
72         }
73
74     cout<<ans<<endl;
75
76     output(cur);
77
78     return 0;
79 }

HDU 1160 FatMouse's Speed 简单DP