[2016-03-27][HDU][1087][Super Jumping! Jumping! Jumping!]

• 时间:2016-03-27 15:51:40 星期日

• 题目编号:[2016-03-27][HDU][1087][Super Jumping! Jumping! Jumping!]

• 分析:dp[i]表示跳到第i个位置,能拿到的最多分,则dp[i] = max(dp[i] , dp[j] + v[i]) 能从j跳到i

• 遇到的问题:排序之和,a[1].v <= a[2].v 而不是 a[1].v < a[2].v 所以还是需要判断一下价值

#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int maxn = 1000 + 10;
LL dp[maxn];
struct pos{
        int v,id;
        pos(int a = -1,int b = -1):v(a),id(b){}
        bool operator < (const pos & a){
                return v < a.v;
        }
}a[maxn];
int main(){
        int n;
        while(~scanf("%d",&n) && n){
                int tmp;
                for(int i = 1;i <= n ;++i){
                        scanf("%d",&tmp);
                        a[i] = pos(tmp,i);
                }
                sort(a+1,a+n+1);
                LL ans = 0;
                for(int i = 1;i <= n ; ++i){
                        dp[i] = a[i].v;
                        for(int j = 1;j < i;++j){
                                if(a[j].id < a[i].id && a[j].v < a[i].v)
                                        dp[i] = max(dp[i] ,dp[j] + a[i].v);
                        }
                        ans = max(ans,dp[i]);
                }
                printf("%I64d\n",ans);
        }
        return 0;
}

来自为知笔记(Wiz)