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
4000
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. 

ps：带权并查集

我们可以把位置坐标分为x轴和y轴，所以有两个权值，x[i]记录i的x轴到其根节点x轴的距离，y[i]记录i的y轴到其根节点y轴的距离，
（压缩路径和合并集合时权值的改变类似于向量的加减）

最后把询问离线存储，再把询问按index排序，之后在询问之前先合并集合，接着再判断就好了

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define maxn 40000+10
#define maxv 10000+10
struct node//存储输入的数据
{
int u,v,length;
char s;
} q[maxn];
struct question//存储询问
{
int u,v,index,pos;
friend bool operator < (question a,question b)
{
return a.index<b.index;
}
} w[maxv];
struct gather//并查集
{
int x,y,pre;
} g[maxn];
struct result//存储答案
{
int ans,pos;
}e[maxv];
int n,m;

bool cmp(result a,result b)
{
return a.pos<b.pos;
}

void init()
{
for(int i=1; i<=n; i++)
{
g[i].pre=i;
g[i].x=g[i].y=0;
}
}

int Find(int x)
{
int temp=g[x].pre;
if(x==g[x].pre)
return x;
g[x].pre=Find(temp);
g[x].x+=g[temp].x;
g[x].y+=g[temp].y;
return g[x].pre;
}

void join(int u,int v,int length,char s)
{
int fx=Find(u),fy=Find(v);
if(fx!=fy)
{
g[fx].pre=fy;
if(s=='N')
g[fx].y=g[v].y+length-g[u].y,g[fx].x=g[v].x-g[u].x;
else if(s=='S')
g[fx].y=g[v].y-length-g[u].y,g[fx].x=g[v].x-g[u].x;
else if(s=='E')
g[fx].x=g[v].x+length-g[u].x,g[fx].y=g[v].y-g[u].y;
else
g[fx].x=g[v].x-length-g[u].x,g[fx].y=g[v].y-g[u].y;
}
}

int main()
{
scanf("%d%d",&n,&m);
init();
int i,k;
for(i=0; i<m; i++)
scanf("%d%d%d %c",&q[i].u,&q[i].v,&q[i].length,&q[i].s);
scanf("%d",&k);
for(i=0; i<k; i++)
scanf("%d%d%d",&w[i].u,&w[i].v,&w[i].index),w[i].pos=i;
sort(w,w+k);
int ans,j=0;
for(i=0; i<k; i++)
{
for(; j<w[i].index; j++)//注意j的值
join(q[j].u,q[j].v,q[j].length,q[j].s);
int u=w[i].u,v=w[i].v;
int fx=Find(u),fy=Find(v);
if(fx!=fy)
ans=-1;
else
ans=abs(g[u].x-g[v].x)+abs(g[u].y-g[v].y);
e[i].ans=ans,e[i].pos=w[i].pos;
}
sort(e,e+k,cmp);
for(int i=0;i<k;++i)//按原顺序输出
printf("%d\n",e[i].ans);
return 0;
}