Problem Descrption

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（棋盤）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

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

問題分析：

玩家從起點開始,最后必須跳到終點.在跳躍過程中,玩家將訪問路徑中的棋子,但每個人都必須從一個棋子跳到另一個絕對更大(你可以假設起點是一個最小值和終點是最大值)。)。所有的球員都不能后退。一個跳躍可以從棋子跳到下一步,也可以穿過許多棋子,甚至你可以直線從起點到終點。當然你在這種情況下是零。一個球員是勝利者當且僅當他能根據他的跳躍方案獲得更大的分數。請注意,您的得分來自您跳轉路徑中的棋子上的值之和。您的任務是根據給定的棋子列表輸出最大值。

#include<iostream> #include<cstring>int main(){int n;int a[1000010],b[1000010],i,j,s;while(scanf("%d",&n)&&n!=0){memset(b,0,sizeof(b));//b數組為遞增序列到此的最大和for(i=1; i<=n; i++)scanf("%d",&a[i]);int k=-30000;//結果初始化為最小值，也有可能是負數for(i=1; i<=n; i++){s=-30000;//初始化為最小值for(j=0; j<i; j++)if(a[j]<a[i]&&s<b[j])s=b[j];//找到a[i]之前的最大遞增和b[i]=s+a[i];//從開始到現在的最大遞增序列和if(b[i]>k)k=b[i];}printf("%d\n",k);}return 0;}

其實就是求最長遞增子序列和

AC代碼：

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