Navigation Nightmare POJ - 1984
Navigation Nightmare
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
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. 題目：回答你一連串牧場之間的距離，在一連串回答中，中間穿插一些問題，問能不能確定x,y之間的距離，可以得話輸出答案，不可以則輸出'-1'。
思路：我們容易想到用座標處理“曼哈頓距離”，帶權並查集有類似向量的性質，剛好和座標可以對應，我們就可以用帶權並查集來處理該問題，
每個點兩個權值，分別表示x座標和y座標，初始化為（0,0），然後四個方向也可以用座標表示，這樣題目就可以解決了。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath> using namespace std; #define ll long long
#define pb push_back
#define fi first
#define se second const int N = 4e4 + ;
struct node
{
int rt, x, y;
}fa[N];
struct info
{
int x, y, d;
char op;
};
struct que
{
int x, y, inx;
};
vector<info > Info;
vector<que > Que;
int n, m, q; int Find(int x){
if(fa[x].rt == x) return x;
else{
int tmp = fa[x].rt;
fa[x].rt = Find(tmp);
fa[x].x += fa[tmp].x;
fa[x].y += fa[tmp].y;
return fa[x].rt;
}
} void Union(int x, int y, int dx, int dy){
int fax = Find(x);
int fay = Find(y); if(fax != fay){
fa[fay].rt = fax;
fa[fay].x = fa[x].x + dx - fa[y].x;
fa[fay].y = fa[x].y + dy - fa[y].y;
}
} void solve()
{
//while(~scanf("%d%d", &n, &m)){
scanf("%d%d", &n, &m);
for(int i = ; i <= n; ++i){
fa[i].rt = i;
fa[i].x = fa[i].y = ;
}
Info.clear();
Que.clear(); int u, v, d, inx;
char op[];
for(int i = ; i <= m; ++i){
scanf("%d%d%d%s", &u, &v, &d, op);
Info.pb({u, v, d, op[]});
}
scanf("%d", &q);
for(int i = ; i <= q; ++i){
scanf("%d%d%d", &u, &v, &inx);
Que.pb({u, v, inx});
}
//sort(Que.begin(), Que.end()); vector<int > ans;
int j = ;
for(int i = ; i < q; ++i){ while(j < Que[i].inx){
int dx = ;
int dy = ;
if(Info[j].op == 'E') dx = Info[j].d;
else if(Info[j].op == 'W') dx = -Info[j].d;
else if(Info[j].op == 'N') dy = Info[j].d;
else if(Info[j].op == 'S') dy = -Info[j].d; Union(Info[j].x, Info[j].y, dx, dy);
j++;
} int fax = Find(Que[i].x);
int fay = Find(Que[i].y);
if(fax != fay) ans.pb(-);
else{
int dx = fa[Que[i].x].x - fa[Que[i].y].x;
int dy = fa[Que[i].x].y - fa[Que[i].y].y;
ans.pb(abs(dx) + abs(dy));
}
} //for(int o = 0; o < l; ++o) printf("ans = %d\n", ans[o]);
for(int o = ; o < q; ++o) printf("%d\n", ans[o]);
//}
} int main()
{ solve(); return ;
}