POJ1984 Navigation Nightmare —— 种类并查集

题目链接：http://poj.org/problem?id=1984

Navigation Nightmare

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 7136		Accepted: 2556
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 #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

 

 

 

 

 

 

题解：

1.由于查询操作还限定了查询时的下标，即可以查询中间状态。所以需要离线处理。

2.普通的种类并查集。只是当前结点与父节点的的相对关系有两个：相对x和相对y。

 

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 1e5+;

 int n, m, k;
 int fa[MAXN], r[MAXN][];   //r[i][0]、r[i][1]分别代表点i的相对横纵坐标

 struct  //保存边的信息
 {
     int u, v, w;
     char dir[];
 }a[MAXN];

 struct node     //保存查询信息
 {
     int u, v, index, id;    //index为查询的下标; id为此查询输入时的下标，用于输出答案
     bool operator<(const node& x)const {    //按查询的下标从小到大排列
         return index<x.index;
     }
 }q[MAXN];
 int ans[MAXN]; //离线操作之保存答案

 int find(int x)
 {
     if(fa[x]==-) return x;
     int pre = find(fa[x]);
     r[x][] += r[fa[x]][]; //累积相对横坐标
     r[x][] += r[fa[x]][]; //累积相对纵坐标
     return fa[x] = pre;
 }

 int Union(int u, int v, int w, char dir)    //当dir为0时， 代表着查询
 {
     //以下为v相对于u的位置
     int xx = , yy = ;
     if(dir=='E') xx = w; if(dir=='W') xx = -w;
     if(dir=='N') yy = w; if(dir=='S') yy = -w;

     int fu = find(u);
     int fv = find(v);
     if(fu==fv)
         return abs(r[u][]-r[v][]) + abs(r[u][]-r[v][]);
     if(dir==) return -;   //如果是查询操作，并且两者不在同一集合，则直接返回-1；

     fa[fv] = fu;
     r[fv][] = -r[v][]+xx+r[u][];
     r[fv][] = -r[v][]+yy+r[u][];
     return -;
 }

 int main()
 {
     scanf("%d%d", &n, &m);
     memset(fa, -, sizeof(fa));
     memset(r, , sizeof());

     for(int i = ; i<=m; i++)
         scanf("%d%d%d%s", &a[i].u, &a[i].v, &a[i].w, a[i].dir);

     scanf("%d", &k);
     for(int i = ; i<=k; i++)
         scanf("%d%d%d", &q[i].u, &q[i].v, &q[i].index), q[i].id = i;
     sort(q+, q++k);   //对查询进行排序

     int t = ;
     for(int i = ; i<=m; i++)
     {
         Union(a[i].u, a[i].v, a[i].w, a[i].dir[]); //合并u 、v
         while(q[t].index==i)    // 是 while 不是 if ！！因为有可能多个询问都在同一个下标。
         {
             ans[q[t].id] = Union(q[t].u, q[t].v, , );
             if(++t>k) break;
         }
     }

     for(int i = ; i<=k; i++)
         printf("%d\n", ans[i]);
 }