HDU 1087 Super Jumping! Jumping! Jumping! (线性dp)
时间：2015-03-26 17:44:21
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Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24915 Accepted Submission(s): 11009
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.
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.
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
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1087
题目大意：求最长上升子序列的和
题目分析：线性dp，dp[i]表示到第i个点时最长上升子序列的和，若val[i] > val[j]则dp[i] = max(dp[i], dp[j] + val[i]) 0 <= j <= i，初始时dp全为1
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const MAX = 1e3 + 5;
ll dp[MAX];
int val[MAX];
int main()
{
int n;
while(scanf("%d", &n) != EOF && n)
{
ll ans = 0;
for(int i = 0; i < n; i++)
{
scanf("%d", &val[i]);
dp[i] = val[i];
}
for(int i = 0; i < n; i++)
for(int j = 0; j <= i; j++)
if(val[i] > val[j])
dp[i] = max(dp[i], dp[j] + val[i]);
for(int i = 0; i < n; i++)
ans = max(ans, dp[i]);
printf("%lld\n", ans);
}
}dp的路好难走，越难越要走
原文：http://blog.csdn.net/tc_to_top/article/details/44651411
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