HDU - 1160 M - FatMouse's Speed

最新推荐文章于 2021-03-22 20:37:17 发布

原创最新推荐文章于 2021-03-22 20:37:17 发布 · 273 阅读

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本篇介绍了一个有趣的问题解决方法：如何通过排序和动态规划找出一系列老鼠，使得它们的体重递增而速度递减，以证明体重与速度之间的特定关系。文章详细解析了算法思路，并附带完整代码实现。

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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：给出一些老鼠的体重和跑速，为了证明老鼠越重跑得越慢，要找一些数据，由若干个老鼠组成，保证老鼠的体重依次增加而跑速依次减小，问这些数据最多能有多少老鼠，并按体重从小到大输出这些老鼠的序号。

分析：建一个结构体记录每只老鼠的体重，速度和序号，然后按照老鼠的体重从小到大排序，然后用动态规划找最长的老鼠速度递减的长度，dp[j]表示前j只老鼠最长下降子序列，找出最大的dp[j],还要用数组记录老鼠的原始序号。

代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1010],f[1010],ans[1010];
struct node
{
    int w,v,d;     //体重  速度  序号
} q[1010];
bool cmp(node a,node b)
{
    if(a.w==b.w)
        return a.v>b.v;
    return a.w<b.w;
}
int main()
{
    int n=0;
    while(~scanf("%d%d",&q[n].w,&q[n].v))
    {
        q[n].d=n+1;
        n++;
    }
    sort(q,q+n,cmp);        //排序
    for(int i=0; i<n; i++)  //初始化
        dp[i]=1;            
    memset(f,0,sizeof(f));  

    for(int i=0; i<n; i++)    //遍历所有的老鼠
        for(int j=0; j<i; j++)
            if(q[i].w>q[j].w&&q[i].v<q[j].v&&dp[i]<dp[j]+1)//状态转移方程
            {
                dp[i]=dp[j]+1;
                f[i]=j;             //记录老鼠原始位置
            }
    int maxn=dp[0],t;
    for(int i=0; i<n; i++)        //找最大的f[i];
        if(dp[i]>maxn)
        {
            maxn=dp[i];
            t=i;
        }
    int v=1;
    printf("%d\n",maxn);
    while(t)
    {
        ans[v++]=t;            //记录老鼠的位置
        t=f[t];
    }
    v--;
    for(int i=v; i>0; i--)
        printf("%d\n",q[ans[i]].d);
}