FatMouse‘s Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13643    Accepted Submission(s): 6011
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

以weight为第一元素从小到大排序，相同时speed为第二元素从大到小，状态转移方程为

dp[i] = 1     dp[i] = max(dp[i],dp[j]+1); (j<i,p[j].weight>p[i].weight,p[j].speed<p[i].speed);

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn = 1005;
 6 struct node{
 7     int weight;
 8     int speed;
 9     int num;
10 };
11 node p[maxn];
12 int dp[maxn];
13 bool cmp(node A,node B){
14     if(A.weight != B.weight) return A.weight<B.weight;
15     else return A.speed>B.speed;
16 }
17 void solve(){
18     int k = 1,x,y;
19     while(scanf("%d%d",&x,&y)!=EOF){
20         p[k].weight = x;
21         p[k].speed = y;
22         p[k].num = k;
23         k++;
24     }
25     sort(p+1,p+k+1,cmp);
26     int ans = 0;
27     for(int i = k;i>=1; i--){
28             dp[i] = 1;
29         for(int j = i; j<=k; j++){
30             if(p[j].weight>p[i].weight&&p[j].speed<p[i].speed){
31                 dp[i] = max(dp[i],dp[j]+1);
32             }
33         }
34         ans = max(ans,dp[i]);
35     }
36     printf("%d\n",ans);
37     for(int i = 1; i<=k; i++){
38         if(dp[i] == ans){
39             printf("%d\n",p[i].num);
40             ans--;
41         }
42         if(ans == 0) break;
43     }
44 }
45 int main()
46 {
47     solve();
48     return 0;
49 }

FatMouse's Speed 基础DP