FatMouse‘s Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9069    Accepted Submission(s):
4022
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

总结：不知道这么题怎么回事，输出的结果和测试数据不一样，但是还是ac了。我的输出结果时4,5,9,8

思路：将数据按照重量递增排序，若重量相等，则按照速度递减排序，本题实则求得是排序后的速度v的最长递减子序列。

dp方程：a[i]=a[j]+1,b[i]=j;数组b用于存储其路径

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<algorithm>
 4 using namespace std;
 5 int b[1001];
 6 int a[1001];
 7 int c[1001];
 8 struct node
 9 {
10     int w;
11     int v;
12     int sign;
13
14 } s[1001];
15 int cmp(node x, node y)
16 {
17     if(x.w==y.w)
18         return x.v>y.v;
19     return x.w<y.w;
20 }
21 int main()
22 {
23     int i,j,flag=-1,k=1,Max=0;
24     while(scanf("%d %d",&s[k].w, &s[k].v)!=EOF)
25     {
26         s[k].sign=k;
27         k++;
28     }
29     sort(s,s+k,cmp);
30     for(i=1; i<k; i++)
31     {
32         a[i]=1;
33         b[i]=-1;
34         for(j=i-1; j>=1; j--)
35         {
36             if(s[i].w>s[j].w && s[i].v<s[j].v && a[j]+1>a[i])
37             {
38                 a[i]=a[j]+1;
39                 b[i]=j;
40             }
41         }
42         if(a[i]>Max)
43         {
44             Max=a[i];
45             flag=i;
46         }
47     }
48     printf("%d\n",Max);
49     int t=0;
50     while(flag!=-1)
51     {
52         c[t]=flag;
53         flag=b[flag];
54         t++;
55     }
56     for(i=t-1; i>=0; i--)
57         printf("%d\n",s[c[i]].sign);
58     return 0;
59 }