zoukankan      html  css  js  c++  java
  • E

    /*

    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. 


    状态分析 node(pos,num,sum) (k+1).sum = max((i--k).sum) +(k+1).num */ #include<iostream> #include<vector> #include<cstdlib> #include<cstdio> using namespace std; int N; int main() { int i,j; int a[1000]; int sum[1000]; cin>>N; while(N!=0) { int all_max = -1; memset(a,-1,sizeof(a)); memset(sum,0,sizeof(sum)); for(i=0;i<N;i++) { cin>>a[i]; if(i==0) { sum[0] = a[i]; continue; } int max = -1; bool sign = true; for(j=0;j<i;j++)//找出i之前位置中小于a[i]的位置中最大的sum,如果没有就sign = true { if(a[j]<a[i]) { sign = false; if(sum[j]>max) max = sum[j]; } } if(sign)//此时sum值为a[i] sum[i] = a[i]; else //否则状态转移方程可知 sum[i] = a[i] + max; if(sum[i]>all_max) all_max = sum[i]; } cout<<all_max<<endl; cin>>N; } }
  • 相关阅读:
    同步linux服务器的时间
    Esper学习之四:Context
    Haproxy+Keepalived搭建Weblogic高可用负载均衡集群
    C++中的对象指针
    第十五周oj刷题——Problem M: C++习题 矩阵求和--重载运算符
    《人工智能教程(张仰森)》(二)
    JAVA 并发编程-线程池(七)
    新手学測试----Unit Test(单元測试)
    HDU2193-AVL-数据结构-AVL
    Visual Assist X破解版安装(vs2010助手)
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6128762.html
Copyright © 2011-2022 走看看