任意门：http://poj.org/problem?id=1984
Navigation Nightmare
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 7783 | Accepted: 2801 | |
| Case Time Limit: 1000MS | ||
Description
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 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.
题意概括：
给出 M 个建树的操作，K 次查询，每次查询 x 到 y 经过前 num 次建树操作的距离，如果未联通则输出-1；
解题思路：
带权并查集，路径压缩采用向量法。
dx【i】表示 i 距离所在树根结点的横坐标
dy【i】表示 i 距离所在树根结点的纵坐标
合并 u v 过程：
先合并两棵子树
ru = getfa（u）
rv = getfa（v）
改变其中一棵子树的根结点
fa[ rv ] = ru;
更新根结点的相对值（之后子树的相对值会通过查找父结点的过程进行更新）
dx[ rv ] = dx[ u ] - dx[ v ] - wx[ u, v];
dy[ rv ] = dy[ u ] - dy[ v ] - wy[ u, v];
先执行num次建树操作
查找父结点的同时压缩路径
查询最后结果
如果相同根，直接计算两点的曼哈顿距离
否则不连通
Tip：
题目没有说查询的 num 是非递减有序的，所以处理查询前要对 num 进行排序！！！
也就是说离线处理，在线处理会出错。（虽然poj上的数据我没有排序也AC了， 但这是需要考虑的情况）
AC code：
//离线带权并查集
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define INF 0x3f3f3f3f
using namespace std;
const int MAXN = 4e4+;
const int MAXK = 1e4+;
struct Query{int x, y, index, no;}q[MAXN]; //查询信息
int fa[MAXN], dx[MAXN], dy[MAXN]; //并查集，路径压缩（距离起点的横坐标距离 和 纵坐标距离）
int u[MAXN], v[MAXN]; //第i个操作的起点 和 终点
int ans[MAXK]; //第 k 次查询的结果
int wx[MAXN], wy[MAXN]; //wx[ i ] 第i个操作的横坐标边权 wy[ i ] 第i个操作的纵坐标边权
int N, M, K; void init()
{
for(int i = ; i <= N; i++){
fa[i] = i;
dx[i] = ;
dy[i] = ;
}
memset(q, , sizeof(q));
}
int aabs(int x){return x>?x:-x;}
int getfa(int s)
{
if(s == fa[s]) return s;
int t = fa[s];
fa[s] = getfa(fa[s]); //压缩路径
dx[s] += dx[t];
dy[s] += dy[t];
return fa[s];
}
bool cmp(Query q1,Query q2){return q1.index < q2.index;}
int main()
{
while(~scanf("%d%d", &N, &M)){
//scanf("%d%d", &N, &M);
init();
char nod;
for(int i = , d; i <= M; i++){
scanf("%d%d%d %c", &u[i], &v[i], &d, &nod);
if(nod == 'E') {wx[i] = d; wy[i] = ;}
else if(nod == 'W') {wx[i] = -d; wy[i] = ;}
else if(nod == 'N') {wy[i] = d; wx[i] = ;}
else if(nod == 'S') {wy[i] = -d; wx[i] = ;}
}
scanf("%d", &K);
for(int i = ; i <= K; i++){
scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].index);
q[i].no = i;
}
sort(q+, q+K+, cmp);
int k = ;
for(int i = ; i <= K; i++){
//printf("i:%d\n", i);
while(k <= q[i].index){ //合并index个操作
//printf("k:%d\n", k);
int ru = getfa(u[k]);
int rv = getfa(v[k]);
//printf("u:%d ru:%d v:%d rv:%d\n", u[k], ru, v[k], rv);
fa[rv] = ru; //合并两个集合
dx[rv] = dx[u[k]] - dx[v[k]] - wx[k]; //
dy[rv] = dy[u[k]] - dy[v[k]] - wy[k];
k++;
}
//printf("k:%d\n", k);
if(getfa(q[i].x) != getfa(q[i].y)) ans[q[i].no] = -; //两点经过index次操作后还是没有相连
else{
ans[q[i].no] = aabs(dx[q[i].x] - dx[q[i].y]) + aabs(dy[q[i].x] - dy[q[i].y]);
}
//puts("");
}
for(int it = ; it <= K; it++) printf("%d\n", ans[it]);
}
return ;
}
一道拖了好久好久的带权并查集，给几个数据纪念一下
Input:
S
S
S
S
N
N
N
Output:
-
-
Input:
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
S
S
S
Output:
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
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