17996 daily cool run (DP)

Time Limit: 1000 ms memory limit: 65535 K
Submission times: 0 pass times: 0

Question type: Programming Language: Not Limited

Description

Daily Cool Run is a popular game, and Xdp enjoys playing the game recently. While playing the game, you may get normal coins or flying coins by running and jumping.Now, he meets a problem that what is the maximum score he can obtain.To simplify the problem, we suppose that the maps of the game are 2 * n retangles, whose second rows are the ground.At the beginning of the game, the player will start from the grid (2, 1) (the lower left corner of a map).During the game, you have two choices, Run or Jump. When you are on the ground of grid (2, i),    1. Run to grid (2, i + 1);    2. Jump to grid (2, i + 3) by go through grids (1, i + 1) and (1, i + 2).Know that you can’t land while jumping , and must follow the path stated above .When you arrive one of the last two grids, the game will be over.Now, Xdp knows the maps of the game, whose grids are assigned to a value x(0 <= x <= 100). If x=0, it means this grid is empty, else it means there is a coin whose value is x.Now, can you tell me what is the maximum score you can get?

Input Format

There are at most 100 cases.The first line is an integer T, the number of the cases.In each case, the first line is a integer n (n <= 10

5

).The Following two lines this a 2 * n rectangular, that means in each line, there are n integers (x1, x2, …. xn).  ( 0 <= x < =100, 0 means that this gird is an empty gird, others represent the coins, x is its value).

Output Format

For each test case, output a single line with an integer indicates the maximum score . 

Input example

280 0 1 1 0 1 1 02 1 0 0 1 0 0 150 0 1 1 02 1 2 0 1

Output example

96
Idea: DP [I] indicates the maximum value obtained when the column I in the second row is obtained in a two-dimensional matrix.
DP [I] = max (DP [I-1] + A [I], DP [I-3] + A [I-2] + A [I-1] + B [I]), I> 3;
DP [I] = DP [I-1] + A [I], I <= 3
Since it will end at the end of a [n] Or B [N], to avoid the trouble of processing, simply extend the map and make the value of the Extended Part 0, this is equivalent to eventually falling onto the ground.

1 # include <cstdio> 2 # include <cstring> 3 # include <algorithm> 4 # include <queue> 5 # include <iostream> 6 # include <cmath> 7 using namespace std; 8 int A [100005], B [100005]; 9 int DP [100005]; 10 int main () 11 {12 INT t; 13 scanf ("% d ", & T); 14 While (t --) 15 {16 int N; 17 scanf ("% d", & N); 18 for (INT I = 1; I <= N; ++ I) scanf ("% d", & A [I]); 19 for (INT I = 1; I <= N; ++ I) scanf ("% d", & B [I]); 20 A [n + 1] = A [n + 2] = A [n + 3] = B [n + 1] = B [n + 2] = B [n + 3] = 0; 21 memset (DP, 0, sizeof DP); 22 for (INT I = 1; I <= N + 3; ++ I) // extend the map length to N + 323 if (I> 3) DP [I] = max (B [I] + A [I-1] + A [I-2] + dp [I-3], B [I] + dp [I-1]); 24 else DP [I] = B [I] + dp [I-1]; 25 printf ("% d \ n", DP [N + 3]); 26} 27}

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17996 daily cool run (DP)