Tokyo 2009 / UVa 1445 Cubist Artwork (二视图中的最少方块)

1445 - Cubist Artwork

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=512&page=show_problem&problem=4191

International Center for Picassonian Cubism

The artwork is a collection of cubes with edges of one foot long and is built on a flat ground that is divided into a grid of one foot by one foot squares. Due to some technical reasons, cubes of the artwork must be either put on the ground, fitting into a unit square in the grid, or put on another cube in the way that the bottom face of the upper cube exactly meets the top face of the lower cube. No other way of putting cubes is possible.

You are a member of the judging committee responsible for selecting one out of a plenty of artwork proposals submitted to the competition. The decision is made primarily based on artistic quality but the cost for installing the artwork is another important factor. Your task is to investigate the installation cost for each proposal. The cost is proportional to the number of cubes, so you have to figure out the minimum number of cubes needed for installation.

Each design proposal of an artwork consists of the front view and the side view (the view seen from the right-hand side), as shown in Figure 1.

Figure 1: An example of an artwork proposal

The front view (resp., the side view) indicates the maximum heights of piles of cubes for each column line (resp., row line) of the grid.

There are several ways to install this proposal of artwork, such as the following figures.

In these figures, the dotted lines on the ground indicate the grid lines. The left figure makes use of 16 cubes, which is not optimal. That is, the artwork can be installed with a fewer number of cubes. Actually, the right one is optimal and only uses 13 cubes. Note that, a single pile of height three in the right figure plays the roles of two such piles in the left one.

Notice that swapping columns of cubes does not change the side view. Similarly, swapping rows does not change the front view. Thus, such swaps do not change the costs of building the artworks. For example, consider the artwork proposal given in Figure 2.

Figure 2: Another example of artwork proposal

An optimal installation of this proposal of artwork can be achieved with 13 cubes, as shown in the following figure, which can be obtained by exchanging the rightmost two columns of the optimal installation of the artwork of Figure 1.

Input

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. Each dataset is formatted as follows.

w d 
h₁ h₂ ... h_w 
h'₁ h'₂ ... h'_d

The integers w and d separated by a space are the numbers of columns and rows of the grid, respectively. You may assume 1w10 and 1d10. The integers separated by a space in the second and third lines specify the shape of the artwork. The integers h_i (1h_i20, 1iw) in the second line give the front view, i.e., the maximum heights of cubes per each column line, ordered from left to right (seen from the front). The integers h'_i (1h'_i20, 1id )

Output

For each dataset, output a line containing the minimum number of cubes. The output should not contain any other extra characters.

You can assume that, for each dataset, there is at least one way to install the artwork.

Sample Input

5 5 1 2 3 4 5 1 2 3 4 5 5 5 2 5 4 1 3 
4 1 5 3 2 
5 5 
1 2 3 4 5 
3 3 3 4 5 
3 3 
7 7 7 
7 7 7 
3 3 
4 4 4 
4 3 4 
4 3 
4 2 2 4 
4 2 1 
4 4 
2 8 8 8 
2 3 8 3 
10 10 
9 9 9 9 9 9 9 9 9 9 
9 9 9 9 9 9 9 9 9 9 
10 9 
20 1 20 20 20 20 20 18 20 20 
20 20 20 20 7 20 20 20 20
0 0

Sample Output

15 15 21 21 15 
13 
32 
90 
186

首先至少得有正视图的正方体的数目。

然后比较正视图和右视图，看是否有一列数目是一样的。

完整代码：

/*0.019s*/

#include<cstdio>
#include<cstring>

int a[11], b[11], vis[21];

int main(void)
{
	int w, d;
	while (scanf("%d%d", &w, &d), w)
	{
		int ans = 0;
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < w; ++i)
		{
			scanf("%d", &a[i]);
			ans += a[i];
			++vis[a[i]];///至少得是这么多
		}
		for (int i = 0; i < d; ++i)
		{
			scanf("%d", &b[i]);
			if (vis[b[i]])
				--vis[b[i]];///比较主视图和右视图一列立方体数相同的个数
			else
				ans += b[i];///主视图没有的话就加上
		}
		printf("%d\n", ans);
	}
	return 0;
}