HDU 1087 Super Jumping! Jumping! Jumping!
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题目链接~~>
做题感悟:这题跟上一题都看了很久,没有忍住百度了一下,然后~~
解题思路:动态方程 F[ i ] = max{ g[ i ] , F[ j ] ( 0<=j<i ) } .
代码:
#include<stdio.h>#include<iostream>#include<map>#include<string>#include<string.h>#include<stdlib.h>#include<queue>#include<algorithm>using namespace std ;const int MX = 300005 ;int g[1005],b[1005] ;int main(){ int n ; while(~scanf("%d",&n)&&n) { for(int i=0 ;i<n ;i++) scanf("%d",&g[i]) ; int best=0 ; b[0]=g[0] ; for(int i=1 ;i<n ;i++) { int mx=g[i] ; for(int j=0 ;j<i ;j++) if(g[j]<g[i]&&b[j]+g[i]>mx) mx=b[j]+g[i] ; b[i]=mx ; best = best > mx ? best : mx ; } printf("%d\n",best) ; } return 0 ;}
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- hdu/hdoj 1087 Super Jumping! Jumping! Jumping!
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