Navigation Nightmare
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 7871 | Accepted: 2831 | |
| Case Time Limit: 1000MS | ||
题目链接：http://poj.org/problem?id=1984
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 #17
There 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 of two farms connected by a road, L is its length, and D is a character that is either 'N', 'E', 'S', or 'W' giving the direction of the road from F1 to F2.
* Line M+2: A single integer, K (1 <= K <= 10,000), the number of FB's queries
* Lines M+3..M+K+2: Each line corresponds to a query from Farmer Bob and contains three space-separated integers: F1, F2, and I. F1 and F2 are numbers of the two farms in the query and I is the index (1 <= I <= M) in the data after which Bob asks the query. 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's queries. Each line should contain either a distance measurement or -1, if it is impossible to determine the appropriate distance.
Sample Input:
7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6 1
1 4 3
2 6 6
Sample Output:
13
-1
10
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.
题意：
给出n个农场，然后按时间依次给出m个关于农场相对位置的信息，之后会给出询问，问在t时刻，x到y的曼哈顿距离是多少。
题解:
这题一开始我以为会用一个时间数组来维护x到y的最大时间，然后直接在线询问进行判断，但发现后来行不通...
之后便发现把输入先储存起来进行离线操作就可以了，具体做法如下：
把询问按照时间从小到大排序，然后按时间对点进行合并，然后用带权并查集维护一下点的x，y值就好了。
更新x，y值可以采用向量法去思考，fx->fy = fx->x + x->y + y->fy ，注意有向性。
代码如下：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <queue> #include <cmath> #include <vector> using namespace std; typedef pair<int,int> pii; const int N = 40005, K = 10005; int n,m,k; int f[N]; struct query{ int p1,p2,t,id; bool operator < (const query &A)const{ return A.t<t; } }q[K]; struct farm{ int X,Y; }p[N]; struct link{ int x,y,dis; char c; }l[N]; int find(int x){ if(x==f[x]) return x; int tmp=f[x]; f[x]=find(f[x]); p[x].X+=p[tmp].X; p[x].Y+=p[tmp].Y; return f[x]; } void Union(int x,int y,int dir,int d){ int fx=find(x),fy=find(y); f[fx]=fy; if(dir==1) p[fx].X=p[y].X-d-p[x].X,p[fx].Y=p[y].Y-p[x].Y;//x在y的西面 else if(dir==2) p[fx].X=p[y].X+d-p[x].X,p[fx].Y=p[y].Y-p[x].Y ; else if(dir==3) p[fx].Y=p[y].Y-d-p[x].Y,p[fx].X=p[y].X-p[x].X;//x在y的南面 else p[fx].Y=d-p[x].Y+p[y].Y,p[fx].X=p[y].X-p[x].X; } int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=m;i++){scanf("%d%d%d %c",&l[i].x,&l[i].y,&l[i].dis,&l[i].c);} scanf("%d",&k); priority_queue <query> que; for(int i=1,x,y,t;i<=k;i++){ scanf("%d%d%d",&q[i].p1,&q[i].p2,&q[i].t);q[i].id=i; que.push(q[i]); } priority_queue <pii ,vector<pii>,greater<pii> > ans ; for(int i=1;i<=N-5;i++) f[i]=i; for(int i=1,x,y,dis,pd;i<=m;i++){ char c; x=l[i].x;y=l[i].y;dis=l[i].dis;c=l[i].c; if(c=='E') pd=1;else if(c=='W') pd=2;else if(c=='N') pd=3;else pd=4; int fx=find(x),fy=find(y); if(fx!=fy) Union(x,y,pd,dis); while(que.top().t==i && !que.empty()){ query now=que.top();que.pop(); int now1=now.p1,now2=now.p2; if(find(now1)==find(now2)){ int xx = abs(p[now1].X-p[now2].X),yy=abs(p[now1].Y-p[now2].Y); ans.push(make_pair(now.id,xx+yy)); }else ans.push(make_pair(now.id,-1)); } } while(!ans.empty()){ printf("%d ",ans.top().second); ans.pop(); } return 0; }