HDU 1160(动态规划-最长上升子序列)

谁借莪１个温暖的怀抱￠ 2022-06-03 07:07 174阅读 0赞

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.

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

题目分析:按照体重排序后，就是求速度的一个最长严格下降的子序列，记录路径

代码如下:

#include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    using namespace std;
    
    const int maxn=1e3+100;
    struct note
    {
        int s,w;
        int cnt;
    }point[maxn];
    int pre[maxn],dp[maxn];
    
    bool cmp(struct note a,struct note b)
    {
        if (a.w==b.w) return a.s>b.s;
        return a.w<b.w;
    }
    void print(int pos)
    {
        if (pre[pos]!=pos) print(pre[pos]);
        printf("%d\n",point[pos].cnt);
    }
    int main()
    {
        int n=0;
        while (scanf("%d%d",&point[n].w,&point[n].s)!=EOF) {
            point[n].cnt=n+1;
            n++;
        }
        sort(point,point+n,cmp);
        memset (dp,0,sizeof (dp));
        int ans=0,pos;
        for (int i=0;i<n;i++) {
            dp[i]=1;
            pre[i]=i;
            for (int j=0;j<i;j++) {
                if (point[i].s<point[j].s&&point[i].w>point[j].w) {
                    if (dp[i]<dp[j]+1) {
                        dp[i]=dp[j]+1;
                        pre[i]=j;//记录路径，它的前一个
                    }
                }
            }
            ans=max(ans,dp[i]);
        }
        for (int i=0;i<n;i++) {
            if (ans==dp[i]) {
                pos=i;
                break;
            }
        }
        printf("%d\n",ans);
        print(pos);//递归输出路径
        return 0;
    }