怒刷DP之 HDU 1160

FatMouse's Speed
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Appoint description: 

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
 
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <queue>
 5 #include <vector>
 6 #include <map>
 7 #include <algorithm>
 8 #include <cstring>
 9 #include <cctype>
10 #include <cstdlib>
11 #include <cmath>
12 #include <ctime>
13 #include <climits>
14 using    namespace    std;
15 
16 const    int    SIZE = 1005;
17 struct    Node
18 {
19     int    pos;
20     int    weight,speed;
21     int    front,num;
22 }DP[SIZE];
23 
24 bool    comp(const Node & r_1,const Node & r_2);
25 int    main(void)
26 {
27     int    count = 0;
28 
29     while(scanf("%d%d",&DP[count].weight,&DP[count].speed) != EOF)
30     {
31         DP[count].pos = count;
32         DP[count].num = 0;
33         count ++;
34     }
35     sort(DP,DP + count,comp);
36 
37     int    max = 1,max_loc = 0;
38     for(int i = 0;i < count;i ++)
39     {
40         DP[i].front = i;
41         for(int j = 0;j < i;j ++)
42             if(DP[i].weight > DP[j].weight && DP[i].speed < DP[j].speed)
43                 if(DP[i].num < DP[j].num)
44                 {
45                     DP[i].num = DP[j].num;
46                     DP[i].front = j;
47                     if(max < DP[i].num + 1)
48                     {
49                         max = DP[i].num + 1;
50                         max_loc = i;
51                     }
52                 }
53         DP[i].num ++;
54     }
55 
56     int    temp[SIZE];
57     count = 0;
58     while(1)
59     {
60         temp[count] = max_loc;
61         if(max_loc == DP[max_loc].front)
62         {
63             count ++;
64             break;
65         }
66         max_loc = DP[max_loc].front;
67         count ++;
68     }
69     printf("%d\n",max);
70     for(int i = count - 1;i >= 0;i --)
71         printf("%d\n",DP[temp[i]].pos + 1);
72 
73     return    0;
74 }
75 
76 bool    comp(const Node & r_1,const Node & r_2)
77 {
78     return    r_1.weight < r_2.weight;
79 }

 

posted @ 2015-09-07 17:21  Decouple  阅读(213)  评论(0)  编辑  收藏  举报