HDU 1160 FatMouse's Speed （动态规划、最长下降子序列）

2022-12-03 20:22:46

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24573 Accepted Submission(s): 10896
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

题目大意与分析

给出一堆鼠标的重量与速度，输出一个最长的序列，满足重量越来越大，速度越来越小。

先将鼠标根据重量从小到大排序，再求最大下降子序列就行了。

#include<bits/stdc++.h>

using namespace std;

typedef struct

{

    int add;

    int w;

    int s;

}node;

node a[];

int dp[],anss,i,temp,c[],x,y,num,j,n;

bool cmp(node a,node b)

{

    return a.w<b.w;

}

int main()

{

    while(cin>>x>>y)

    {

        num++;

        a[num].add=num;

        a[num].w=x;

        a[num].s=y;

    }

    sort(a+,a++num,cmp);

    for(i=;i<=num;i++)

    {

        dp[i]=;

        for(j=;j<i;j++)

        {

            if(a[j].w<a[i].w&&a[j].s>a[i].s)

            dp[i]=max(dp[j]+,dp[i]);

        }

        if(dp[i]>anss)

        anss=dp[i];

    }

    cout<<anss<<endl;

    temp=anss;

    for(i=num;i>=;i--)

    {

        if(dp[i]==temp)

        {

            temp--;

            c[temp]=a[i].add;

        }

    }

    for(i=;i<anss;i++)

    cout<<c[i]<<endl;

}

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