HDU 1160 FatMouse's Speed （dp）

最新推荐文章于 2021-06-01 18:00:21 发布

原创最新推荐文章于 2021-06-01 18:00:21 发布 · 316 阅读

0 ·

CC 4.0 BY-SA版权

普通 dp 专栏收录该内容

4 篇文章

订阅专栏

解决FatMouse的速度挑战问题，寻找体重递增且速度递减的最大子序列，并输出该子序列的索引。通过排序和动态规划算法实现。

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18264 Accepted Submission(s): 8057
Special Judge

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

Sample Output

题意要求尽可能找体重上升的并且速度下降的例子，并且输出路径，所以可以按照体重降序排序然后找最多体重下降时速度上升的倒叙输出路径即可

记忆路径方法:比如记忆下标分别为 1 4 5 7 2 的路径可使a[4]=1,a[5]=4,a[7]=5,a[2]=7;

#include <stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node{
    int speed;
    int weight;
    int num;
}a[1001];
int cmp(node a,node b)
{
    return a.weight>b.weight;
}
int main(int argc, char *argv[])
{
    int cnt=0;
    while(scanf("%d %d",&a[cnt].weight,&a[cnt].speed)!=EOF)
    {
        a[cnt].num=cnt+1;
        cnt++;
    }
    int i,j;
    int dp[1001];
    for(int i=0;i<cnt;i++)dp[i]=1;
    int path[1001];
    sort(a,a+cnt,cmp);
    memset(path,-1,sizeof(path));
    for(i=0;i<cnt;i++)
    {
        for(j=0;j<i;j++)
        {
            if(a[i].speed>a[j].speed&&dp[i]<dp[j]+1&&a[i].weight<a[j].weight)//取a[i] 
            {
                 dp[i]=dp[j]+1;
                path[i]=j;
            }
        }
    }
    int max1=-1;
    int pos=0;
    for(i=0;i<cnt;i++)
    {
        if(max1<dp[i])
        {
            max1=dp[i];
            pos=i;
        }
    }
    printf("%d\n",max1);
    while(pos!=-1)
    {
        printf("%d\n",a[pos].num);
        pos=path[pos];
    }
    return 0;
}