poj1984 带权并查集（向量处理）

Navigation Nightmare

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 5939		Accepted: 2102
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. 

 

题意:

有n个地点，m条边，对于每条边，有信息x,y,len,d,其中len表示x,y间的长度，d表示y相对于x的位置。现在有q个查询，每个查询有x,y,I,其中I表示完成I条路的时候当前2个点的状态。

 

思路:

首先很容易想到这事一题带权并查集的题目，因为各个点之间都有关系。但是关键在于如何处理这里的位置。现在我们用rel[x][0]表示x点到其所在树的父亲节点的向量的横坐标，rel[x][1]表示x点到其所在树的父亲节点的向量的纵坐标 ，其余的操作就是普通的带权并查集了。不过要注意一下这里的更新的操作（因为是向量）和每条边x,y的关系（更新时候谁当父节点）。

 

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<time.h>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = 40010;
struct node
{
    int x;
    int y;
    int z;
    int id;
    char s[10];
    ll ans;
}q[10010],a[MAXN];
int pa[MAXN],n,m,k,siz[MAXN];
ll rel[MAXN][2];
void Init()
{
    for(int i = 0; i <= n; i++){
        pa[i] = i;
        siz[i] = 1;
    }
    memset(rel,0,sizeof(rel));
}
bool cmp1(node fa,node fb)
{
    return fa.z < fb.z;
}
bool cmp2(node fa,node fb)
{
    return fa.id < fb.id;
}
void getp(int& x,int& y,node fp,int flag)
{
    if(fp.s[0] == 'E'){
        x = fp.z;
        y = 0;
    }
    else if(fp.s[0] == 'W'){
        x = - fp.z;
        y = 0;
    }
    else if(fp.s[0] == 'S'){
        x = 0;
        y = fp.z;
    }
    else {
        x = 0;
        y = - fp.z;
    }
    if(flag)x *= -1, y *= -1;
}
int find(int x)
{
    if(x != pa[x]){
        int fx = find(pa[x]);
        siz[fx] += siz[x];
        rel[x][0] = rel[x][0] + rel[pa[x]][0];
        rel[x][1] = rel[x][1] + rel[pa[x]][1];
        pa[x] = fx;
    }
    return pa[x];
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        Init();
        for(int i = 0; i < m; i++){
            scanf("%d%d%d%s",&a[i].x,&a[i].y,&a[i].z,a[i].s);
        }
        scanf("%d",&k);
        for(int i = 0; i < k; i++){
            q[i].id = i;
            scanf("%d%d%d",&q[i].x,&q[i].y,&q[i].z);
        }
        sort(q,q+k,cmp1);
        int p = 0;
        for(int i = 0; i < m; i++){
            int x = a[i].x;
            int y = a[i].y;
            int tx = 0,ty = 0;
            int fx = find(x);
            int fy = find(y);
            if(fx != fy){
                if(siz[fx] > 1){
                    getp(tx,ty,a[i],0);
                    siz[fx] += siz[fy];
                    pa[fy] = fx;
                    rel[fy][0] = -rel[y][0] - tx + rel[x][0];
                    rel[fy][1] = -rel[y][1] - ty + rel[x][1];
                }
                else {
                    getp(tx,ty,a[i],1);
                    siz[fy] += siz[fx];
                    pa[fx] = fy;
                    rel[fx][0] = rel[y][0] - tx - rel[x][0];
                    rel[fx][1] = rel[y][1] - ty - rel[x][1];
                }
            }
            while(p < k && q[p].z == i + 1){
                x = q[p].x;
                y = q[p].y;
                fx = find(x);
                fy = find(y);
                if(fx != fy){
                    q[p].ans = -1;
                }
                else {
                   // cout<<rel[x][0]<<' '<<rel[y][0]<<' '<<rel[x][1]<<' '<<rel[y][1]<<endl;
                    q[p].ans = fabs(rel[x][0] - rel[y][0]) + fabs(rel[x][1] - rel[y][1]);
                }
                p ++;
            }
        }
        sort(q,q+k,cmp2);
        for(int i = 0; i < k; i++){
            printf("%lld
",q[i].ans);
        }
    }
    return 0;
}