Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50936 Accepted Submission(s): 23645
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
题意 给出一行棋子，他们有各自的权值，要求找出权值递增的一个子串（不一定连续但顺序不能变） 这个子串权值相加和最大，输出这个最大和
状态转移方程：ans[k]=max(a[k]+ans[j],ans[k]) j<k a[j]<a[k]
1 #include<bits/stdc++.h> 2 using namespace std; 3 int a[1005]; 4 long long ans[1005]; 5 int main() 6 { 7 int n; 8 while(~scanf("%d",&n)&n) 9 { 10 memset(a,0,sizeof(a)); 11 for(int i=1; i<=n; i++) 12 { 13 scanf("%d",&a[i]); 14 } 15 memset(ans,0,sizeof(ans)); 16 long long maxx=0; 17 for(int i=1;i<=n;i++) 18 { 19 for(int j=0;j<i;j++) 20 { 21 if(a[j]<a[i]&&ans[j]+a[i]>ans[i])//因为i前面的每一个数都和i比较 a[j]<a[i]计算到i为止的 22 { //和时，肯定会加上a[i] 所以算和的时候也是目前i位置上的和 与目前j位置上 23 ans[i]=ans[j]+a[i];// 的和加上a[i]比较 ，如果目前i位置上的和小，就要更新一下 24 if(ans[i]>maxx)maxx=ans[i];//每次都判断一下，最后得到最大值 25 } 26 } 27 } 28 29 printf("%lld\n",maxx); 30 } 31 return 0; 32 }