HDU - 1087 Super Jumping! Jumping! Jumping!(dp)
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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\n", ans); } return 0; }
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