HDU

题意：从起点依次跳跃带有数字的点直到终点，要求跳跃点上的数字严格递增，问跳跃点的最大数字和。

分析：

1、若之前的点比该点数字小，则可进行状态转移，dp[i] = max(dp[i], dp[j] + a[i]);

2、dp[i]---截止到i，跳跃的最大数字和。

3、由于不确定最终是哪个点直接跳往终点可保证数字和最大，因此，扫一遍，ans = max(ans, dp[i]);

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
int dp[MAXN];
int main(){
    int N;
    while(scanf("%d", &N) == 1){
        if(N == 0) return 0;
        memset(dp, 0, sizeof dp);
        for(int i = 1; i <= N; ++i){
            scanf("%d", &a[i]);
        }
        for(int i = 1; i <= N; ++i){
            dp[i] = a[i];
            for(int j = 1; j <= i - 1; ++j){
                if(a[j] < a[i]){
                    dp[i] = max(dp[i], dp[j] + a[i]);
                }
            }
        }
        int ans = 0;
        for(int i = 1; i <= N; ++i){
            ans = max(ans, dp[i]);
        }
        printf("%d
", ans);
    }
    return 0;
}