FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 177 Accepted Submission(s): 90

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 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

Sample Output

分析：

首先按照关键字W升序进行排序，然后S[i]从1到i-1(是指外层循环中的第i个元素)开始遍历，

找到w[i]>w[k],s[i]

动态规划状态方程为：f[i] = max(f[k]+1,f[i]),1<=k

这里定义了一个struct mice 其中的len代表当前元素的状态长度，用来找到找到最长的，

index用来记忆当前元素的真实下标，在排序后还能找到元素的下标。

before用来找的当前元素的上一个元素的下标。

#include <iostream>

#include <stdlib.h>

using namespace std;
typedef struct
{

    int w,s;
    int len,index;
    int before;
}mice;
int cmp(const void *a,const void *b)
{

mice c = *(mice *)a;

    mice d = *(mice *)b;
    if(c.w==d.w)
    return d.s - c.s;
    else return c.w - d.w;
}
int main()
{

    mice m[10001];
    int i = 1,flag,max=0,f[1001];
    while(cin>>m[i].w>>m[i].s)
    {

m[i].index = i;

m[i].len=0;

m[i].before=0;

i++;
}

//cout<<i<<endl;

    qsort(m,i-1,sizeof(m[1]),cmp);
    for(int j = 1;j < i ;j++)
    {

        for(int k = 1; k< j;k++)
        {

            if(m[j].w>m[k].w&&m[j].s<m[k].s)
            {

                if(m[j].len<m[k].len+1)
                {

m[j].len = m[k].len+1;

                    m[j].before = k;
                    if(m[j].len>=max)
                    {

max = m[j].len;

                        flag = j;
                    }
                }
            }
        }
    }

cout<<max+1<<endl;

f[1] = m[flag].index;

    i=2;
     while(m[flag].before!=0)
    {

flag = m[flag].before;

f[i] = m[flag].index;

        i++;
    }
    for(int j = i-1 ; j >=1 ; j--)
    {

cout<<f[j]<<endl;
}

// for(int j = 1 ; j < i ;j++)

// cout<<m[j].index<<" "<<m[j].w<<" "<<m[j].s<<endl;

return 0;
}

本文转自NewPanderKing51CTO博客，原文链接：

http://www.cnblogs.com/newpanderking/archive/2011/08/04/2126977.html

，如需转载请自行联系原作者