题目链接：http://poj.org/problem?id=1984
Time Limit: 2000MS Memory Limit: 30000K 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：
有n个节点，给出m条数据，每条数据包含F1,F2,L,D，代表从节点F1到F2的距离为L，D为'E/W/S/N'，代表了F1→F2是指向东/西/南/北；
现在又有个人来给出k条询问，每条询问包含F1,F2,IDX，代表了查询节点F1和F2之间的距离，本次查询发生在录入第IDX条数据之后（也就是说本次查询时，第IDX+1条往后的数据都还是未知的）；
注意：对于查询的回答，必须按照查询输入的顺序进行输出；同时可能在读入第idx条数据之后，读入第idx+1条数据之前，会有多个查询。
题解：
并查集建树，par[x]代表x的父亲节点，val[x].X和val[x].Y分别代表par[x]->x向量的水平分量和竖直分量；
注意做好find()函数内val[x]的更新、unite两个节点时val[]更新，并且注意将答案按照查询的顺序输出即可。
AC代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=+; int n,m,k; int par[maxn];
struct Val{
int X,Y; //par[x]->x向量的水平分量和竖直分量
}val[maxn];
void init(int l,int r){for(int i=l;i<=r;i++) par[i]=i,val[i].X=val[i].Y=;}
int find(int x)
{
if(par[x]==x) return x;
else
{
int root=find(par[x]);
val[x].X+=val[par[x]].X;
val[x].Y+=val[par[x]].Y;
return par[x]=root;
}
} struct Data{
int F1,F2,L;
char D[];
}data[maxn]; vector<int> D2Q[maxn]; //Data->Query
struct Query{
int F1,F2;
int id; //记录下是第id个查询
}query[maxn]; struct Res{
int val; //第id个查询的答案值
int id; //代表本结果是对应到第id个查询的
Res(int val,int id){this->val=val,this->id=id;}
bool operator<(const Res &oth)const
{
return id<oth.id;
}
}; int main()
{
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++) scanf("%d%d%d%s",&data[i].F1,&data[i].F2,&data[i].L,data[i].D); scanf("%d",&k);
for(int i=;i<=n;i++) D2Q[i].clear();
for(int i=,idx;i<=k;i++)
{
scanf("%d%d%d",&query[i].F1,&query[i].F2,&idx);
query[i].id=i;
D2Q[idx].push_back(query[i].id); //记录一下第i个询问发生在第idx个数据之后
} init(,n);
vector<Res> res;
for(int i=,a,b,t1,t2;i<=m;i++)
{
a=data[i].F1, b=data[i].F2;
t1=find(a), t2=find(b);
if(t1!=t2)
{
par[t2]=t1;
int dX,dY; //dX是a->b向量的水平分量，dY是a->b向量的竖直分量
if(data[i].D[]=='E') dX=data[i].L, dY=;
if(data[i].D[]=='W') dX=-data[i].L, dY=;
if(data[i].D[]=='N') dX=, dY=data[i].L;
if(data[i].D[]=='S') dX=, dY=-data[i].L;
val[t2].X=val[a].X+dX-val[b].X;
val[t2].Y=val[a].Y+dY-val[b].Y;
} //在录入本次数据之后，查看是否有查询，若有尝试进行回答
for(int j=,_size=D2Q[i].size();j<_size;j++)
{
Query Q=query[D2Q[i][j]];
a=Q.F1, b=Q.F2;
t1=find(a), t2=find(b); int ans;
if(t1!=t2) ans=-;
else ans=abs(val[a].X-val[b].X)+abs(val[a].Y-val[b].Y); res.push_back(Res(ans,Q.id)); //将查询结果进行记录
}
} sort(res.begin(),res.end()); //将查询的结果按照之前查询的顺序排列
for(int i=;i<res.size();i++) printf("%d\n",res[i].val);
}