Navigation NightmareDescription
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
|
F7Being 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 6Sample Output
13
-1
10Hint
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
到 N。牧场间存在一些道路，每条道路道路连接两个不同的牧场，方向必定平行于 X 轴或 Y 轴。乡
下地方的道路不会太多，连通两座牧场之间的路径是唯一的。
突然间，约翰的导航仪失灵了，牧场的坐标记录全部消失了。所幸的是，约翰找到了表示道路的
数据，可以通过这些信息得知牧场间的相对位置。但贝西有急事，在约翰工作到一半的时候就要知道
一些牧场间的曼哈顿距离。这时，如果约翰能从找回的道路信息之间推算出答案，就会告诉贝西。请
你帮助约翰来回答贝西的问题吧。 ( x 1 , y 1) 和 ( x 2 , y 2) 间的曼哈顿距离定义为 | x 1 − x 2| + | y1 − y2|。
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define Maxn 40010 int fa[Maxn],nx[Maxn],ny[Maxn]; int myabs(int x) {return x>?x:-x;} int ffind(int x)
{
int y=fa[x];
if(x!=fa[x]) fa[x]=ffind(fa[x]);
nx[x]+=nx[y];ny[x]+=ny[y];
return fa[x];
} char s[]; struct node
{
int x,y,c,p;
int ans;
}t[],tt[]; bool cmp(node x,node y) {return x.c<y.c;}
bool cmp2(node x,node y) {return x.p<y.p;} int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++) fa[i]=i,nx[i]=ny[i]=;
for(int i=;i<=m;i++)
{
int x,y,c;
scanf("%d%d%d%s",&x,&y,&c,s); t[i].x=x;t[i].y=y;t[i].c=c;
if(s[]=='E') t[i].p=;
else if(s[]=='W') t[i].p=;
else if(s[]=='N') t[i].p=;
else t[i].p=;
}
int q;
scanf("%d",&q);
for(int i=;i<=q;i++)
{
int x,y,c;
scanf("%d%d%d",&x,&y,&c);
tt[i].x=x;tt[i].y=y;tt[i].c=c;
tt[i].p=i;
}
sort(tt+,tt++q,cmp);
int now=;
for(int i=;i<=q;i++)
{
while(now<tt[i].c)
{
now++;
int x=t[now].x,y=t[now].y,c=t[now].c;
int ff=ffind(y);
nx[ff]=-nx[y];ny[ff]=-ny[y];fa[ff]=y;
nx[y]=ny[y]=;fa[y]=x; if(t[now].p==) nx[y]=c;
else if(t[now].p==) nx[y]=-c;
else if(t[now].p==) ny[y]=c;
else ny[y]=-c;
} int x=tt[i].x,y=tt[i].y;
if(ffind(x)!=ffind(y)) tt[i].ans=-;
else
{
tt[i].ans=myabs(nx[x]-nx[y])+myabs(ny[x]-ny[y]);
}
}
sort(tt+,tt++q,cmp2);
for(int i=;i<=q;i++) printf("%d\n",tt[i].ans);
return ;
}
[POJ 1984]
2016-10-27 18:26:37