hdu 1087 Super Jumping! Jumping! Jumping!
2017-02-18 00:25
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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.
InputInput 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.
OutputFor each case, print the maximum according to rules, and one line one case.
Sample Input
Sample Output
题意:求取最大(不是最长)递增子段和;注意数据:5 1 6 3 4 5 结果:13;
思路:算是郑重的接触dp了吧,不过有点坎坷了;dp [ i ]表示以a [ i ]为最大值的递增子段和,有点递推的意思,详见代码;
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.
InputInput 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.
OutputFor 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:求取最大(不是最长)递增子段和;注意数据:5 1 6 3 4 5 结果:13;
思路:算是郑重的接触dp了吧,不过有点坎坷了;dp [ i ]表示以a [ i ]为最大值的递增子段和,有点递推的意思,详见代码;
#include<cstdio> #include<algorithm> #include<cmath> using namespace std; int a[1010],dp[1010]; int main() { int n,i,j; while(scanf("%d",&n) && n) { for(i=0;i<n;i++) scanf("%d",&a[i]); int maxn=0,ans; for(i=0;i<n;i++) { dp[i]=a[i]; for(j=0;j<i;j++)//蒟蒻博主卡在了(j=i+1;j<n;j++); { ans=a[i]; if(a[i]>a[j]) ans+=dp[j];//a[i]结尾,和dp[i]连接; dp[i]=max(ans,dp[i]);//选取以a[i]结尾的最大dp; } maxn=max(maxn,dp[i]);//取最终的最大值,也就是输出结果; } printf("%d\n",maxn); } return 0; }
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