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  • hdu1160最长递减子序列及其路径

    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
    ",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
    ",s[c[i]].sign);
    58     return 0;
    59 }
    View Code
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  • 原文地址:https://www.cnblogs.com/lxm940130740/p/3871099.html
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