首页 > ACM题库 > HDU-杭电 > hdu 1933 Ballroom Lights待解决[解题报告]C++
2013
12-26

hdu 1933 Ballroom Lights待解决[解题报告]C++

Ballroom Lights

问题描述 :

The ICPC world finals will be held in a luxurious hotel with a big ballroom. A buffet meal will be served in this ballroom, and organizers decided to decorate its walls with pictures of past champion teams.

In order to avoid criticism about favouring some of those teams over others, the organizing commitee wants to make sure that all pictures are appropiately illuminated.

The only direct way they’ve found for doing this is ensuring each picture has at least one lightbulb that directlyilluminates it.

In this way, the perimeter of the ballroom wall can be divided into illuminated parts (in whichpictures may be placed) and dark parts (which are not suitable for placing the pictures).

The ballroom has the shape of a box and contains several lightbulbs. Each lightbulb emits light in all directions, but this light can be blocked by columns. All columns in the ballroom have cylindrical shape and go from the floor to the ceiling, so light cannot pass over or above them. Columns are of course placed so that its circular section is parallel to the ballroom floor.

Any given point p on the perimeter wall is said to be illuminated if there exists a line segment (a light ray) which starts on a lightbulb, ends in p and does not touch or pass through any column.

Your task as a helper of the ICPC organization is to examine the blueprints of the ballroom and determine the total length of illuminated sections of the perimeter wall. The blueprint consist of a rectangle indicating a top view of the ballroom, with the lightbulbs and columns marked in it.

输入:

Each test case will consist on several lines. The first line will contain four integers: L, the number of lightbulbs, C, the number of columns, X, the size of the ballroom on the x coordinate and Y , the size of the ballroom on the y coordinate. The lower-left corner of the ballroom is at (0, 0) while the upper-right corner is at (X, Y ).

The next L lines will contain two integers each representing the x and y coordinate of each lightbulb. The last C lines of the test case will contain three integers each, representing the x and y coordinates of the center of a column and its radius, in that order. You can assume that 1 <= L,C <= 103 and 4 <= X, Y <= 106. Also, for all pairs of coordinates (x,y), 0 < x < X and 0 < y < Y , both for lightbulbs and column center locations. All radii of the columns will be positive. Finally, no two columns will overlap, although they may touch, and no column will touch or intersect with the border of the ballroom. No lightbulb will be inside a column or in its boundary and no two lightbulbs will be in the same place.
Input is terminated with L = C = X = Y = 0.

输出:

Each test case will consist on several lines. The first line will contain four integers: L, the number of lightbulbs, C, the number of columns, X, the size of the ballroom on the x coordinate and Y , the size of the ballroom on the y coordinate. The lower-left corner of the ballroom is at (0, 0) while the upper-right corner is at (X, Y ).

The next L lines will contain two integers each representing the x and y coordinate of each lightbulb. The last C lines of the test case will contain three integers each, representing the x and y coordinates of the center of a column and its radius, in that order. You can assume that 1 <= L,C <= 103 and 4 <= X, Y <= 106. Also, for all pairs of coordinates (x,y), 0 < x < X and 0 < y < Y , both for lightbulbs and column center locations. All radii of the columns will be positive. Finally, no two columns will overlap, although they may touch, and no column will touch or intersect with the border of the ballroom. No lightbulb will be inside a column or in its boundary and no two lightbulbs will be in the same place.
Input is terminated with L = C = X = Y = 0.

样例输入:

2 1 8 8
6 6
2 6
4 4 2
1 4 7 7
3 3
2 4 1
4 2 1
2 2 1
4 4 1
2 2 9 7
1 2
5 5
3 3 2
7 5 1
0 0 0 0

样例输出:

28.0000
0.0000
25.8214


  1. 给你一组数据吧:29 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1000。此时的数据量还是很小的,耗时却不短。这种方法确实可以,当然或许还有其他的优化方案,但是优化只能针对某些数据,不太可能在所有情况下都能在可接受的时间内求解出答案。

  2. 第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。

  3. int half(int *array,int len,int key)
    {
    int l=0,r=len;
    while(l<r)
    {
    int m=(l+r)>>1;
    if(key>array )l=m+1;
    else if(key<array )r=m;
    else return m;
    }
    return -1;
    }
    这种就能避免一些Bug
    l,m,r
    左边是l,m;右边就是m+1,r;