FatMouse’s Speed#
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10172 Accepted Submission(s): 4521 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
先排序，然后求一个最长上升子序列。
第一次做记录路径的题，比较无力。
调了很久，后来调到样例都不对整个人都无奈了。
可能也是调了太久的缘故，整个人比较烦躁，也忘记了题目中的那句"如果有多组解那么任意输出一种"，结果白白浪费了好久。sad
还是不要浮躁。
AC代码：
代码实现
1
2
3
4 /* ***********************************************
5 Author :111qqz
6 Created Time :2016年02月22日 星期一 23时23分41秒
7 File Name :code/hdu/1160.cpp
8 ************************************************ */
9
10 #include <iostream>
11 #include <algorithm>
12 #include <cstring>
13 #include <cstdio>
14 #include <cmath>
15
16 using namespace std;
17
18 int n,tmp,tmppre,m,index;
19 const int N=1E5+5;
20 int dp[N],ans[N],pre[N],k;
21
22 struct ST
23 {
24 int w,s,id;
25 }st[N];
26
27 bool cmp(ST a,ST b)
28 {
29 if ( a.w<b.w) return true;
30 if ( a.w==b.w&&a.s>b.s ) return true;
31 return false;
32 }
33 int main()
34 {
35 n = 1;
36 while(scanf("%d %d",&st[n].w,&st[n].s)!=EOF)
37 {
38
39 st[n].id = n;
40 n++;
41 // if (n>9) break;
42
43 }
44 n--;
45 memset(dp,0,sizeof(dp));
46 memset(ans,0,sizeof(ans));
47 memset(pre,-1,sizeof(pre));zixulie
48 sort(st+1,st+n+1,cmp);
49
50
51 dp[1] = 1;
52 for ( int i = 2; i <= n; i++)
53 {
54 dp[i] = 1;
55 for ( int j = 1 ; j <= i -1 ; j++ )
56 if (st[j].w<st[i].w&&st[j].s>st[i].s&&dp[j]+1>dp[i])
57 {
58
59 dp[i] = dp[j]+1;
60 pre[i] = j;
61 }
62 }
63 // cout<<endl;
64 // for ( int i = 1 ; i <= n ; i++ )
65 // cout<<st[i].w<<" "<<st[i].s<<endl;
66 // for ( int i = 1 ; i <= n ; i++ )
67 // cout<<"pre[i]:"<<pre[st[i].id]<<"dp[i]:"<<dp[i]<<endl;
68
69
70 m = -1;
71 for ( int i = 1 ; i <= n;i++ )
72 if ( dp[i] > m)
73 {
74 m = dp[i];
75 index = i;
76 }
77 cout<<m<<endl;
78 // cout<<"index "<<index<<endl;
79 k = 0;
80 while (index!=-1)
81 {
82 k++;
83 ans[k] = index;
84 index=pre[index];
85 }
86 for ( int i = k; i >= 1; i-- )
87 cout<<st[ans[i]].id<<endl;
88
89
90 return 0;
91 }