Navigation Nightmare
Time Limit: 2000MSMemory Limit: 30000K
Total Submissions: 4599Accepted: 1734
Case Time Limit: 1000MS
Description
Farmer John’s pastoral neighborhood has N farms (2 <= N <= 40,000), usually numbered/labeled 1..N. A series of M (1 <= M < 40,000) vertical and horizontal roads each of varying lengths (1 <= length <= 1000) connect the farms. A map of these farms might look something like the illustration below in which farms are labeled F1..F7 for clarity and lengths between connected farms are shown as (n):F1 — (13) —- F6 — (9) —– F3||(3)||(7)F4 — (20) ——– F2|||(2)F5|F7 Being an ASCII diagram, it is not precisely to scale, of course.Each farm can connect directly to at most four other farms via roads that lead exactly north, south, east, and/or west. Moreover, farms are only located at the endpoints of roads, and some farm can be found at every endpoint of every road. No two roads cross, and precisely one path (sequence of roads) links every pair of farms.FJ lost his paper copy of the farm map and he wants to reconstruct it from backup information on his computer. This data contains lines like the following, one for every road:There is a road of length 10 running north from Farm #23 to Farm #17There is a road of length 7 running east from Farm #1 to Farm #17…As FJ is retrieving this data, he is occasionally interrupted by questions such as the following that he receives from his navigationally-challenged neighbor, farmer Bob:What is the Manhattan distance between farms #1 and #23?FJ answers Bob, when he can (sometimes he doesn’t yet have enough data yet). In the example above, the answer would be 17, since Bob wants to know the "Manhattan" distance between the pair of farms.The Manhattan distance between two points (x1,y1) and (x2,y2) is just |x1-x2| + |y1-y2| (which is the distance a taxicab in a large city must travel over city streets in a perfect grid to connect two x,y points).When Bob asks about a particular pair of farms, FJ might not yet have enough information to deduce the distance between them; in this case, FJ apologizes profusely and replies with "-1".
Input
* Line 1: Two space-separated integers: N and M* Lines 2..M+1: Each line contains four space-separated entities, F1,F2, L, and D that describe a road. F1 and F2 are numbers oftwo farms connected by a road, L is its length, and D is acharacter that is either ‘N’, ‘E’, ‘S’, or ‘W’ giving thedirection of the road from F1 to F2.* Line M+2: A single integer, K (1 <= K <= 10,000), the number of FB’squeries* Lines M+3..M+K+2: Each line corresponds to a query from Farmer Boband contains three space-separated integers: F1, F2, and I. F1and F2 are numbers of the two farms in the query and I is theindex (1 <= I <= M) in the data after which Bob asks thequery. Data index 1 is on line 2 of the input data, and so on.
Output
* Lines 1..K: One integer per line, the response to each of Bob’squeries. Each line should contain either a distancemeasurement or -1, if it is impossible to determine theappropriate distance.
Sample Input
7 61 6 13 E6 3 9 E3 5 7 S4 1 3 N2 4 20 W4 7 2 S31 6 11 4 32 6 6
Sample Output
13-110
Hint
At time 1, FJ knows the distance between 1 and 6 is 13.At time 3, the distance between 1 and 4 is still unknown.At the end, location 6 is 3 units west and 7 north of 2, so the distance is 10.
Source
题目链接：?id=1984题目大意：在一个图中，只有上下左右四个方向的边。给出这样的一些边，求任意2个节点之间的曼哈顿距离。题目分析：代权并查集，由于要根据时间来判断当前状态，又题目说了查询时间是递增的，我们不妨先把输入的信息存储下来，根据查询的时间来将信息加入集合中，设定两个数组x[i],y[i]分别表示i点相对于原点（这里的原点即我们设定的根结点）的横坐标和纵坐标，方向就同二维坐标系一样，通过方向来改变坐标值，压缩路径时要更新w[i]的值，合并的过程和CD题相同，采用向量法。r1->r2 = r1->a + a->b + b->r2，因为w[i]表示i对原点的偏移量，所以w[r2] = w[a] + d -w[b]，若两点在同一集合中直接计算出其曼哈顿距离，若两点不在同一集合中，说明不连通，则输出－1模拟一遍样例，这里是执行完Find(1-7)以后的坐标t = 1: x[1] = 0， y[1] = 0 x[6] = 13， y[6] = 0t = 2: x[1] = 0， y[1] = 0 x[6] = 13， y[6] = 0 x[3] = 22， y[3] = 0t = 3: x[1] = 0， y[1] = 0 x[6] = 13， y[6] = 0 x[3] = 22， y[3] = 0 x[5] = 22， y[5] = -7t = 4(换原点)： x[4] = 0， y[4] = 0 x[1] = 0， y[1] = 3 x[6] = 13， y[6] = 3 x[3] = 22， y[3] = 3 x[5] = 22， y[5] = -4t = 5(换原点)： x[2] = 0， y[2] = 0 x[4] = -20，y[4] = 0 x[1] = -20，y[1] = 3 x[6] = -7，， y[6] = 3 x[3] = 2， y[3] = 3 x[5] = 2， y[5] = -4t = 6: x[2] = 0， y[2] = 0 x[4] = -20，y[4] = 0 x[1] = -20，y[1] = 3 x[6] = -7， y[6] = 3 x[3] = 2， y[3] = 3 x[5] = 2， y[5] = -4 x[7] = -20，y[7] = -2而只有在充满了艰辛的人生旅途中，
