FatMouse's Speed

标签：end   ams   for   sort   man   seq   bsp   class   content   

InputInput 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. 
OutputYour 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

Sponsor

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 struct node{
 7     int m,v,num;
 8 }e[1001];
 9 bool cmp(node a,node b){
10     if(a.m==b.m) return a.v>b.v;
11     return a.m<b.m;
12 } 
13 int dp[1010],a[1010];
14 int main()
15 {
16     int i,j,k=1,s=0,t;
17     while(~scanf("%d%d",&e[k].m,&e[k].v))
18     {
19         e[k].num=k++;
20     }
21     sort(e+1,e+1+k,cmp);
22     for(i=1;i<=k;i++)
23     {
24         dp[i]=1;
25         for(j=1;j<i;j++)
26         {
27             if(e[j].m<e[i].m&&e[j].v>e[i].v)
28             dp[i]=max(dp[i],dp[j]+1);
29         }
30         if(dp[i]>s)
31         s=dp[i]; 
32     }
33     t=s;
34     for(i=k-1;i>=0;i--)
35     {
36         if(dp[i]==t){
37             t--;
38             a[t]=e[i].num;
39         }
40     }
41     printf("%d\n",s);
42     for(i=0;i<s;i++)
43     printf("%d\n",a[i]);
44 }

FatMouse's Speed

标签：end   ams   for   sort   man   seq   bsp   class   content   

原文地址：https://www.cnblogs.com/csx-zzh/p/13708820.html