hdu4723 How Long Do You Have to Draw 贪心

How Long Do You Have to Draw

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 322    Accepted Submission(s): 128

Problem Description

There are two horizontal lines on the XoY plane. One is y₁ = a, the other is y₂ = b(a < b). On line y₁, there are N points from left to right, the x-coordinate of which are x = c₁, c₂, ... , c_N (c₁ < c₂ < ... < c_N) respectively. And there are also M points on line y₂ from left to right. The x-coordinate of the M points are x = d₁, d₂, ... d_M (d₁ < d₂ < ... < d_M) respectively.
Now you can draw segments between the points on y₁ and y₂ by some segments. Each segment should exactly connect one point on y₁ with one point on y₂.
The segments cannot cross with each other. By doing so, these segments, along with y1 and y2, can form some triangles, which have positive areas and have no segments inside them.
The problem is, to get as much triangles as possible, what is the minimum sum of the length of these segments you draw?

 

Input

The first line has a number T (T <= 20) , indicating the number of test cases.
For each test case, first line has two numbers a and b (0 <= a, b <= 10⁴), which is the position of y₁ and y₂.
The second line has two numbers N and M (1 <= N, M <= 10⁵), which is the number of points on y₁ and y₂.
The third line has N numbers c₁, c₂, .... , c_N(0 <= c_i < c_i+1 <= 10⁶), which is the x-coordinate of the N points on line y₁.
The fourth line has M numbers d₁, d₂, ... , d_M(0 <= d_i < d_i+1 <= 10⁶), which is the x-coordinate of the M points on line y₂.

 

Output

For test case X, output "Case #X: " first, then output one number, rounded to 0.01, as the minimum total length of the segments you draw.

 

Sample Input

1
0 1
2 3
1 3
0 2 4

 

Sample Output

Case #1: 5.66

 

既然先要满足三角形最多，那么肯定是要把所有的点能连的都连起来，然后重点是最短线段长度，如图

黑线是已经连接的线，x和y是当前位置，下一条线有两种连接方法，即绿线和蓝线，c[x]->d[y + 1]和 d[y]->c[x + 1]，因为c[x]->d[y + 1]较短，所以选这条，y到y + 1的位置上去，然后把这条线段累加上，如果蓝线较短的话则将x放到x + 1的位置上去，然后把这条线段累加上，注意初始化和最后的边界，每一步都这样择优，最后输出的线段长度和一定最小

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>

using namespace std;

double c[100005], d[100005];

int main()
{
	int T, n, m, x, y;
	double a, b, h, len;
	scanf("%d", &T);
	for (int t = 1; t <= T; t++) {
		scanf("%lf%lf", &a, &b);
		h = fabs(a - b) * fabs(a - b);
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++)
			scanf("%lf", &c[i]);
		for (int i = 0; i < m; i++)
			scanf("%lf", &d[i]);
		x = y = 0;
		len = 0;
		while (x < n && y < m) {
			len += sqrt(h + fabs(c[x] - d[y]) * fabs(c[x] - d[y]));
			if (x == n - 1)
				y++;
			else if (y == m - 1)
				x++;
			else {
				if (fabs(c[x] - d[y + 1]) < fabs(c[x + 1] - d[y]))
					y++;
				else
					x++;
			}
		}
		printf("Case #%d: %.2f\n", t, len);
	}
	return 0;
}