HDOJ 1087 Super Jumping! Jumping! Jumping! (DP)
2017-07-10 21:01
309 查看
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.
![](http://acm.hdu.edu.cn/data/images/1087-1.jpg)
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.
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
题目大意:输入n,然后n个数,代表棋盘上除终点和起点外还有n个得分点,要求从起点开始,每一步只能跳到分数比当前点a高的点b,同时获得b分。起点视为无限小,终点视为无限大,可以直接从起点跳到终点,但是得分为零,求跳到终点最高的得分。
题解:从起点开始遍历每个点,更新当前点可以跳到的点的得分,转移方程为dp[j]=max(dp[j],dp[i]+a[i])。
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.
![](http://acm.hdu.edu.cn/data/images/1087-1.jpg)
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.
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
题目大意:输入n,然后n个数,代表棋盘上除终点和起点外还有n个得分点,要求从起点开始,每一步只能跳到分数比当前点a高的点b,同时获得b分。起点视为无限小,终点视为无限大,可以直接从起点跳到终点,但是得分为零,求跳到终点最高的得分。
题解:从起点开始遍历每个点,更新当前点可以跳到的点的得分,转移方程为dp[j]=max(dp[j],dp[i]+a[i])。
#include<stdio.h> #include<algorithm> #include<string.h> using namespace std; long long dp[1005]; int a[1005]; int main() { int n; while(scanf("%d", &n),n) { for(int i=1;i<=n;i++)scanf("%d",&a[i]); for(int i=1;i<=n;i++){ for(int j=i+1;j<=(n+1);j++){ if(a[i]<a[j]||j==(n+1)){ dp[j]=max(dp[j],dp[i]+a[i]); } } } printf("%lld\n",dp[n+1]); memset(dp,0,sizeof(dp)); } return 0; }
相关文章推荐
- HDOJ 1087 Super Jumping! Jumping! Jumping! (dp)
- hdoj 1087Super Jumping! Jumping! Jumping!【dp】
- hdoj Super Jumping! Jumping! Jumping! 1087 (DP求单调递增最大值)
- HDOJ/HDU 1087 Super Jumping! Jumping! Jumping!(经典DP~)
- HDOJ_1087_Super Jumping! Jumping! Jumping! 【DP】
- HDOJ-----1087Super Jumping! Jumping! Jumping!(DP)
- hdoj1087 Super Jumping! Jumping! Jumping! ( dp )
- HDOJ/HDU 1087 Super Jumping! Jumping! Jumping!(经典DP~)
- HDOJ 1087 Super Jumping! Jumping! Jumping!简单DP
- hdoj1087Super Jumping! Jumping! Jumping!【dp】
- hdoj 1087 Super Jumping! Jumping! Jumping! 【dp&&最大递增子段和】
- hdoj 1087Super Jumping! Jumping! Jumping!《《dp》》
- HDOJ 1087 Super Jumping! (DP)
- HDOJ 1087-Super Jumping! Jumping! Jumping!【DP】
- hdoj1087 Super Jumping! Jumping! Jumping!(DP)
- HDU 1087 Super Jumping! Jumping! Jumping! DP
- hdu 1087 Super Jumping! Jumping! Jumping!(水DP)
- hdoj--1087--Super Jumping! Jumping! Jumping!(最长上升序列)
- HDU-1087 Super Jumping! Jumping! Jumping! (DP)
- HDOJ1087 Super Jumping! Jumping! Jumping!【逆推】----武科大ACM暑期集训队选拔赛6题