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 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.
For each case, print the maximum according to rules, and one line one case.
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
4
10
3
题意：在给出的数字序列中，找出一个子序列 满足（1）顺序上升 （2）上升序列求和最大
思路：最长上升子序列改一下就好了
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std ;
#define maxn 2000
int num[maxn] ;
int dp[maxn] ;
int main(){
int n ;
while(~scanf("%d" , &n) && n ){
memset(dp , 0 , sizeof(dp)) ;
for(int i=1 ; i<=n ; i++){
scanf("%d" , &num[i]) ;
dp[i] = num[i] ;
// 初始（子）状态 每一个单独的数字 都可以看成一个 上升子序列
}
int max_num = dp[1] ;
for(int i=1 ; i<=n ; i++){
for(int j=1 ; j<i ; j++){
if(num[j] < num[i]){
dp[i] = max(dp[i] , dp[j] + num[i]) ;
if(dp[i] > max_num){
max_num = dp[i] ;
}
}
}
}
printf("%d\n" , max_num ) ;
}
return 0 ;
}
