First, consider the case of 1 * n. Figure: n = even number in Figure 4 below: n = odd 5
For the FIG., 1 and consider the four corners of 3.1 with at least two lamps. 3 of the four corners there are at least two lamps, and the lamp 1 and 3 share is impossible, because there is no common point 13, 13 so want to satisfy the condition, a minimum of four lamps. May wish to make light appear in the picture on the local loop, I found that if such a lamp is installed, only 13 meet the conditions, the entire 1 * n squares conditions are met. Promotion can be obtained, 1 * n (n is an even number) minimum lamp is n
For the FIG., 1,3,5 consideration. Because it is impossible to share 1,3,5 lamp, so we want to satisfy the condition 135, a minimum of six lights. May wish to make light appear in the picture on the local loop, I found that if such a lamp is installed, only 135 meet the conditions, the entire 1 * n squares conditions are met. Promotion can be obtained, 1 * n (n is an odd number) of the lamp is least n + 1
Consider n, m are both the case of an odd number. For row 3, the two rows no common vertices so that the lamp can not be shared. For the first line, you want to make the bank to meet the meaning of the questions, requires a minimum of six lights, empathy, for 3 lines, also requires six lights. So for the entire plaza, it takes at least 12 lights. When the lamp position as shown in FIG circle lamp 12 exactly, and may satisfy a condition so that the square, so that the minimum number of lamp 12. By promoting, can be properly n, when m is an odd number are required minimum lamp (m + 1) (n + 1) / 2.
Consider the case where n is an even number m is an odd number. For row 3, the two rows no common vertices so that the lamp can not be shared. For the first line, you want to make the bank to meet the meaning of the questions, requires a minimum of four lights, empathy, for 2 lines, also requires four lights. So for the entire plaza, it takes at least eight lights. When the lamp position as shown in FIG circle, exactly eight lights, and may satisfy a condition so that the square, so that the minimum number of lamp 8. By promoting, can be properly n is an even number m is an odd number, the required minimum lamp n (m + 1) / 2.
For n, m all even when, not so simple. If further by previous methods, the first three considered alone, will give the least number of light> = 4. But the actual number is the minimum lamp 5. The reason is that 4 is an even number, 4/2 is an integer, which determines according to the original idea to consider some of these squares no two lamps will shine.
That rectangle on the map for 2 * 4, the minimum number of lamp 5 is how to get it? Suppose the brightness of a light generated in a block where is 1, a, brightness of the lamp e at the generated are 2, b, c, brightness d generated four large rectangular four lamps at the apex brightness 1. For the FIG., A total of eight blocks, the minimum brightness of the lamp 16 is required to be generated. First consider the first column, b must be at the lamp. Because the point b "feature" Up, "radiation" widest range, so b must be used. Below 8 * 10 analysis about the process:
Consider first the upper left corner (1,1) box. This block has at least two points of the four corners of the lamp. I would like to prove that Italy meet minimal installation of lamp posts installed at the midpoint of 1,1 policy certainly lamp: 1,1 points assume no lights, so a total of four cases, namely 0,0 point, the point 0, 1, a lamp points 1,0; 0,0 point, the lamp points 0,1; 1,0 point, the lamp points 0,1; 0,0 point, the lamp points 1,0; 0,0 because only according to a box, with no advantage other than the two points. These three points and the full lighting no other meaning but more than a lamp, it may be the only point 1,0, 0,1 point there is light. We reintroduction 1,0 1,1 0,1,1,1 bright light in both cases. Program 1,0 1,1 1,0 compared with 0,1, 0,1 and 1,1 point is different. Point 0,1 1,1 1,2 light allows blocks, points 1,1 1,1 1,2 2,1 allows blocks 2,2 light. So 1,1 points "feature" completely include the point 0,1 "functional." So scheme 1,0 1,1 1,0 0,1 superior program. Similarly scheme 0,1 1,1 0,1 1,0 superior program. 1,0 1,1 0,1 1,1 but can not compare with the quality. Here we analyze these two programs.
0,1 、1,0 ：
For the 2,1 block, its four vertices of the lamp there must be two. Because there are 1,1 point lights, so to judge the remaining three points in what is certainly a lamp. Because the point 2,1 features up to and including the functions of other points, it is the most advantage of 2,1 points, 2,1 points so definitely lights. Similarly point 2,2, 2,3, 2,4 ??? 2,10 definitely have lights.
Next analysis block 3 and the like through the analysis of previously found, block 3 there are two options, one is the other point is a point 0,3,1,3 1,2, 1,3. We assume program points 1, 1,3 superior points 0,3,1,3 . The next analysis block 3,3, 3,3 for a block, the program of the candidate point and a little 3,2 2,3,3,3 3,3 2,3,3,3 better point is assumed. And then analysis block 4,3. 4,3 lighting points available will continue analysis block 5,3, ··· ··· 10,3 5,3 10,3 available point will be lit. Readily by observation program points 1, 1,3 0,3,1,3 difference in point, the "known" conflict (this is known not really known, but we assume that known). So assumption does not hold, the point 3,2, 3,3 better, empathy, points 5,2 5,3 better, point 7,2,7,3 9,2 points, 9,3 better. Then go to the right, with the same determination method, the last available in the choice of the final point of the program when 0,1 1,1 1,2 1,3 below.
Similarly available at selected points 0,1 1,1 1,3, the final solution below 0,3 time.
This can be obtained in both cases the minimum number of lamp selection points 0,1,1,1. ,
1,0 1,1 same protocol analysis dots 0,1 1,1 analysis, after obtaining the minimum number of points 1,0 1,1 lamp, the smaller value between the two is the final answer. Extended to n * mn, m are even. Minimum number of lights min (n, m) / 2 * max (n, m) +1
In fact, it can be assumed that at the start point 0,1 1,1 1,0 1,1 points better than can be obtained directly , saving the middle of a big step. However, in order to facilitate their understanding, speak intermediate steps are possible.