| 시간 제한 | 메모리 제한 | 제출 | 정답 | 맞힌 사람 | 정답 비율 |
|---|---|---|---|---|---|
| 4 초 (추가 시간 없음) | 1024 MB (추가 메모리 없음) | 40 | 9 | 6 | 35.294% |
You are given a regular hexagon having sides of length $N$. A regular hexagon can be split into unit equilateral triangles of side length 1ドル$ as shown in the figure below. We are going to completely fill the hexagon with unit rhombuses of side length 1ドル$ formed by joining two equilateral triangles which share an edge.
Hexagon formed from triangles
For each position a unit rhombus can be placed, the cost of placing a rhombus is given. Find the minimum cost required to fill the hexagon.
The first line of input contains $N$.
The following 2ドルN$ lines contain the cost for a rhombus placed in each respective row.
Let’s say the cost of a rhombus formed by joining the $j$-th and $j+1$-th triangles of the $i$-th row is $p_{i,j}$.
The $i$-th of the 2ドルN$ lines of input contains $p_{i,1},p_{i,2},\ldots$.
The next 2ドルN-1$ lines of input contain the cost for a rhombus placed across two rows.
Let’s say the cost of a rhombus formed by joining the $j$-th inverted triangle of the $i$-th row and the triangle above it is $q_{i,j}$.
The $i$-th of the 2ドルN-1$ lines contains $q_{i+1,1},q_{i+1,2},\ldots$.
Print the minimum cost required to fill the hexagon using unit rhombuses. It can be proved that it is always possible to fill a hexagon using unit rhombuses.
1 2 3 4 5 1 6
9
2 3 14 15 9 2 6 5 3 5 8 97 9 3 2 3 8 4 6 26 4 3 3 8 3 2 7 9 5 0 2
58
The costs of rhombuses given in example 1
The solution for example 2
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