首页 > ACM题库 > HDU-杭电 > HDU 1160 FatMouse’s Speed-动态规划-[解题报告] C++
2013
12-03

HDU 1160 FatMouse’s Speed-动态规划-[解题报告] C++

FatMouse’s Speed

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

6008 1300
6000 2100
500 2000
1000 4000
1100 3000
6000 2000
8000 1400
6000 1200
2000 1900

样例输出:

4
4
5
9
7

点击打开链接

1,把问题转换为最长上升子序列来做,先把体重或速度排序,然后状态转移方程是dp[i]=max(dp[j])+1,其中(0<j<i && w[i]>w[j] && s[i]<s[j]) 。其中dp[i]表示以i为底的最长序列。

2,怎么存储序列点,用index[i]存储i之前的那个点的坐标。

3,因为在排序过程中要打乱数列,所以把位置也放在结构体中。

#include"stdio.h"
#include"stdlib.h"
#include"string.h"
struct node
{
    int n,w,s;
}mice[1111];
int dp[1111];
int index[1111],a[1111];
int cmp(const void*a,const void*b)
{
    return (*(struct node*)a).w-(*(struct node*)b).w;
}
int main()
{
    int lv=1;
    int i,j,k,max;
    memset(index,-1,sizeof(index));
    while(scanf("%d%d",&mice[lv].w,&mice[lv].s)!=EOF)
    {
        mice[lv].n=lv;
        lv++;
    }
    qsort(mice,lv,sizeof(mice[0]),cmp);
    for(i=0;i<=lv;i++)
        dp[i]=1;
    for(i=2;i<lv;i++)
    {
        for(j=1;j<i;j++)
        {
            if(mice[i].s<mice[j].s)
            {
                if(dp[j]+1>dp[i])
                {
                    dp[i]=dp[j]+1;
                    index[mice[i].n]=mice[j].n;
                }
            }
        }
    }
	
    for(max=-1,i=1;i<=lv;i++)
    {
        if(dp[i]>max)
        {
            max=dp[i];
            k=mice[i].n;
        }
    }
    printf("%d\n",max);
    for(i=max;i>=1;i--)
    {
        a[i]=k;
        k=index[k];
    }
    for(i=1;i<=max;i++)
        printf("%d\n",a[i]);
    return 0;
}


  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. A猴子认识的所有猴子和B猴子认识的所有猴子都能认识,这句话用《爱屋及乌》描述比较容易理解……

  3. 学算法中的数据结构学到一定程度会乐此不疲的,比如其中的2-3树,类似的红黑树,我甚至可以自己写个逻辑文件系统结构来。