Navigation Nightmare---poj1984（多关系并查集）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <map>
#include <vector>
using namespace std;
typedef long long LL;
#define PI 4*atan(1.0)
#define N 42000
#define met(a, b) memset(a, b, sizeof(a))

int f[N], rx[N], ry[N], ans[N];
///rx[i]表示i和f[i]的x的偏移量，ry[i]是y的偏移量；
struct node
{
    int u, v, d, I, Id;
    char dir[10];
}a[N], q[N];

int cmp(node p, node q)
{
    return p.Id<q.Id;
}

int Find(int x)
{
    int k = f[x];
    if(x!=f[x])
    {
        f[x] = Find(f[x]);
        rx[x] += rx[k];
        ry[x] += ry[k];
    }
    return f[x];
}

void Union(int num)
{
    int x = a[num].u, y = a[num].v;
    char ch = a[num].dir[0];

    int px = Find(x), py = Find(y);

    if(px != py)
    {
        f[px] = py;

        rx[px] = rx[y] - rx[x];
        ry[px] = ry[y] - ry[x];

        if(ch == ‘E‘)
            rx[px] -= a[num].d ;
        if(ch == ‘W‘)
            rx[px] += a[num].d;
        if(ch == ‘N‘)
            ry[px] -= a[num].d;
        if(ch == ‘S‘)
            ry[px] += a[num].d;
    }
}
int main()
{
    int n, m, Q;
    while(scanf("%d %d", &n, &m)!=EOF)
    {
        met(a, 0);met(q, 0); met(ans, 0);
        for(int i=0; i<=n; i++)
            f[i] = i, rx[i] = ry[i] = 0;

        for(int i=1; i<=m; i++)
            scanf("%d %d %d %s", &a[i].u, &a[i].v, &a[i].d, a[i].dir);

        scanf("%d", &Q);
        for(int i=1; i<=Q; i++)
        {
            scanf("%d %d %d", &q[i].u, &q[i].v, &q[i].I);
            q[i].Id = i;
        }

        sort(q+1, q+Q+1, cmp);///不排序也可以；因为是按I的顺序给的；
        for(int pre=0, i=1; i<=Q; i++)
        {
            for(int j=pre+1; j<=q[i].I; j++)
                Union(j);

            pre = q[i].I;
            int u = q[i].u, v = q[i].v;

            int px = Find(u);
            int py = Find(v);

            if(px != py) ans[q[i].Id] = -1;
            else ans[q[i].Id] = abs(rx[u]-rx[v]) + abs(ry[u]-ry[v]);
        }
        for(int i=1; i<=Q; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}