【POJ 1984】Navigation Nightmare（带权并查集）

Navigation Nightmare

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. 

Source

USACO 2004 February

 

 

【题意】

　　约翰所在的乡村可以看做一个二维平面，其中有 N 座牧场，每座牧场都有自己的坐标，编号为 1
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|。

 

【分析】

　　妈妈啊这题打的一大半都是搞这个输入，在线就在线，要不要这么恶心！！！醉~~

　　啊，带权并查集，就是还原并查集为一棵树，然后在改变fa的时候改一下就好了。

　　这棵并查集树当然是层数越少越好咯。权值我记录的是它的横坐标与纵坐标跟父亲的差值，更改父亲的时候（就是抛弃原来的父亲的时候）把父亲的权值加到自己身上再修改。

　　然后合并两棵树，就先把一个点弄成那个并查集树的跟，其实做一遍find就已经把它放在根的儿子了，再交换一下根就好了，然后就把整颗树插进去就好了。

　　嗯，自己YY一下都可以Y出来的啦。。

　　代码长度绝对是因为输入太！！恶心！！

 

代码如下：

 #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