Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 63733 Accepted Submission(s): 29669
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
Author
lcy
题意：
一条直线上有N个点，每个点有对应值，从最左边的起点（视为0）开始，跳到右边值大于它的点上，然后将值更新为跳到的点的值，问跳到终点（视为无限大）所经过的值的和最大为多少。
题解：
由题目最多只有一千个点可以联想到直接DP即可，每个点枚举前面所有的点找出能跳到它的点的和最大的点（我在说什么……）。即用dp[N]记录每个点的最大的和。具体见下代码；
#define _CRT_SECURE_NO_DepRECATE #define _CRT_SECURE_NO_WARNINGS #include <cstdio> #include <iostream> #include <cmath> #include <iomanip> #include <string> #include <algorithm> #include <bitset> #include <cstdlib> #include <cctype> #include <iterator> #include <vector> #include <cstring> #include <cassert> #include <map> #include <queue> #include <set> #include <stack> #define ll long long #define INF 0x3f3f3f3f #define ld long double const ld pi = acos(-1.0L), eps = 1e-8; int qx[4] = { 0,0,1,-1 }, qy[4] = { 1,-1,0,0 }, qxx[2] = { 1,-1 }, qyy[2] = { 1,-1 }; using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0); ll n, num[1001], dp[1001], maxx; while (cin >> n && n) { memset(dp, 0, sizeof(dp)); for (int i = 0; i < n; i++) { cin >> num[i]; } for (int i = 0; i < n; i++) { maxx = 0; for (int f = 0; f < i; f++)//遍历找出能跳到i的最大的值 { if (num[i] > num[f]) { maxx = max(dp[f], maxx); } } dp[i] = maxx + num[i]; } for (int i = 0; i < n; i++) { maxx = max(maxx, dp[i]); } cout << maxx << endl; } return 0; }