HDU 1087 Super Jumping! Jumping! Jumping!（求LSI序列元素的和，改一下LIS转移方程）

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1087

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47055    Accepted Submission(s): 21755

Problem Description

Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.



The game can be played by two or more than two players. It consists of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
Your task is to output the maximum value according to the given chessmen list.

 

 

Input

Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.

 

 

Output

For each case, print the maximum according to rules, and one line one case.

 

 

Sample Input

3 1 3 2 4 1 2 3 4 4 3 3 2 1 0

 

 

Sample Output

4 10 3

 

 

Author

lcy

分析：

普通的LIS问题，但是题目不是要求LIS序列的长度，而是LIS序列元素的值

改一下DP方程就ok

dp[0]=a[0]

dp[i]=max(dp[j]+a[i])  a[j]<a[i]且j<i

代码如下：

#include<cstring>
#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        int a[n];
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int dp[n];
        dp[0]=a[0];
        for(int i=1;i<n;i++)
        {
            int max_v=0;
            for(int j=0;j<i;j++)
            {
                if(a[j]<a[i])
                {
                    if(max_v<dp[j])
                    {
                        max_v=dp[j];
                    }
                }
            }
            dp[i]=max_v+a[i];
        }
        int max_v=dp[0];
        for(int i=1;i<n;i++)
        {
            if(max_v<dp[i])
            {
                max_v=dp[i];
            }
        }
        printf("%d\n",max_v);
    }
    return 0;
}

 回来更新一下：

今天训练赛又遇到了这个题目

贴一下代码（比一开始精简了点，因为理解了）

#include<bits/stdc++.h>
using namespace std;
#define INF 99999
int main()
{
    //最大递增子序列和
    int n;
    while(cin>>n,n)
    {
        int a[n];
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int dp[n];
        memset(dp,0,sizeof(dp));
        dp[0]=a[0];
        int ans=0;
        for(int i=1;i<n;i++)
        {
            int maxx=0;
            for(int j=0;j<i;j++)
            {
                if(a[j]<a[i])
                {
                    maxx=max(dp[j],maxx);
                }
            }
            dp[i]=maxx+a[i];
            ans=max(ans,dp[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}