HDU 1087 Super Jumping! Jumping! Jumping! (DP)

原题传送门

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题面

Super Jumping! Jumping! Jumping!

   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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.

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

题面描述

从起点开始跳跃到终点,可以直接跳过去,也可以中途停在某个点上,但起点值必定要比停下来的点的值要小(等于也不能).求出所停下的点的和的最大值.

题目分析

动态规划.枚举每一个点出发,尽可能地在能停下来的点停下.有dp[end] = max ( dp[end] , dp[start]+value[end] );然后查找每一个点作为终点前最后一个点时的值中最大的那一个就是答案了.

具体代码

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <queue>
#include <string.h>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
const int maxn = 1005 ;
int arr[maxn] ;
int DP[maxn] ;
int main()
{
    int n ;
    while(scanf("%d", &n ), n ){
        for( int i = 1 ; i <= n ; i++ ){
            scanf("%d", &arr[i] ) ;
            DP[i] = arr[i] ;//从起点跳到点i,再直接跳到终点
        }
        for( int i = 1 ; i <= n ; i++ ){
            for( int j = 1 ; j < i ; j++ ){
                if( arr[i] > arr[j] && DP[i] < DP[j]+arr[i] ){//下一个点j的value必须大于点i的value
                    DP[i] = DP[j]+arr[i] ;//目标点本身的值小于新的值才更新
                }
            }
        }
        int ans = 0 ;
        for( int i = 1 ; i <= n ; i++ ){
            ans = max( ans , DP[i] ) ;
        }
        printf("%d\n", ans ) ;
    }
    return 0;
}