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(pán)）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(wèn)題連接：http://acm.hdu.edu.cn/showproblem.php?pid=1087

問(wèn)題分析：

玩家從起點(diǎn)開(kāi)始,最后必須跳到終點(diǎn).在跳躍過(guò)程中,玩家將訪(fǎng)問(wèn)路徑中的棋子,但每個(gè)人都必須從一個(gè)棋子跳到另一個(gè)絕對(duì)更大(你可以假設(shè)起點(diǎn)是一個(gè)最小值和終點(diǎn)是最大值)。)。所有的球員都不能后退。一個(gè)跳躍可以從棋子跳到下一步,也可以穿過(guò)許多棋子,甚至你可以直線(xiàn)從起點(diǎn)到終點(diǎn)。當(dāng)然你在這種情況下是零。一個(gè)球員是勝利者當(dāng)且僅當(dāng)他能根據(jù)他的跳躍方案獲得更大的分?jǐn)?shù)。請(qǐng)注意,您的得分來(lái)自您跳轉(zhuǎn)路徑中的棋子上的值之和。您的任務(wù)是根據(jù)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(shù)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(jié)果初始化為最小值，也有可能是負(fù)數(shù)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(kāi)始到現(xiàn)在的最大遞增序列和if(b[i]>k)k=b[i];}printf("%d\n",k);}return 0;}

其實(shí)就是求最長(zhǎng)遞增子序列和

AC代碼：

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