（HDU - 1160）FatMouse's Speed（DP）

最新推荐文章于 2023-03-01 11:20:18 发布

feng_zhiyu

最新推荐文章于 2023-03-01 11:20:18 发布

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分类专栏： HDU DP 文章标签： dp

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本文介绍了一个有趣的问题：如何找出一系列老鼠中体重递增而速度递减的最长序列，并给出了详细的算法实现过程，包括数据结构设计、排序与动态规划的应用。

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Time limit1000 msMemory limit32768 kBOSWindows

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

题意：输入一些老鼠的体重和速度，求体重递增并且速度递减的最长序列的长度并输出是哪几只老鼠？
分析：要求体重递增速度递减，就先按体重从小到大排序，然后采用DP，dp[i]表示在第i只老鼠的最长序列长度，这里要求输出是哪几只老鼠，也就是编号，用一个pre数组记录dp更新时的对应老鼠的下标，最后利用pre记录的下标反序输出

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
#define mem(a,n) memset(a,n,sizeof(a))
const int N=1e3+5;
int dp[N],ans[N];
int n,pre[N];
struct Node
{
    int w,s;
    int xb;//记录老鼠的下标
    bool operator < (const Node& a)const
    {
        return w<a.w;///从小到大排序
    }
} m[N];
int main()
{
    int cnt=0;
    while(scanf("%d%d",&m[cnt].w,&m[cnt].s)!=EOF)
    {
        m[cnt].xb=cnt;
        cnt++;
    }
    mem(pre,-1);///初始化
    sort(m,m+cnt);
    int k=0;
    for(int i=0; i<cnt; i++)
    {
        dp[i]=1;///至少为1
        for(int j=0; j<i; j++)
        {
            if(m[i].w>m[j].w&&m[i].s<m[j].s&&dp[i]<dp[j]+1)///满足体重递增，速度递减且长度可以更新
            {
                dp[i]=dp[j]+1;
                pre[m[i].xb]=m[j].xb;///记录当前老鼠对应的下一只老鼠的下标
            }
        }
        if(dp[i]>dp[k]) k=i;///k始终为最长序列的长度对应的下标
    }
    int MAX=dp[k],i=0;
    printf("%d\n",MAX);
    k=m[k].xb;
    while(k!=-1)///终止条件
    {
        ans[i++]=k;///ans记录输出的下标
        //printf("k=%d\n",k);
        k=pre[k];
    }
    for(int i=MAX-1; i>=0; i--)///反序输出
        printf("%d\n",ans[i]+1);
    return 0;
}