POJ 1984 Navigation Nightmare (并查集)

题意：给定一组线段和方向，然后查询点的距离。

思路：并查集的基本操作，在记录坐标偏移的时候注意一下，两个点之间和他们的根之间的坐标偏移关系可以用关系式表达出来，只要在纸上写一写就ok。查询输入的时候还是按ind排下序在处理线段。

/*
 * 简单题 并查集的使用，就当是复习了。
 */
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 40005;
const int INF = 1000000000;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

typedef struct{
    int x,y;
}Node;
typedef struct{
    int a,b;
    int x,y;
}Road;
typedef struct{
    int a,b;
    int ind;
}Ask;
bool cmp(const Ask& a,const Ask& b)
{
    return a.ind < b.ind;
}

Node node[MAXN];
Road road[MAXN*1000];
Ask qus[MAXN*10];
int n,m,k;
int pre[MAXN];

int init()
{
    for(int i = 0; i < MAXN; ++i)
        node[i].x = node[i].y = 0;
    CLR(pre,-1);
    return 0;
}
int find_set(int x)
{
    int root = x;
    int c_px,c_py,p_py,p_px;
    int tmp = 0;
    p_px = node[x].x;
    p_py = node[x].y;
    while(pre[root] >= 0)
    {
        root = pre[root];
        node[x].x += node[root].x;
        node[x].y += node[root].y;
    }
    while( x != root)
    {
        tmp = pre[x];
        c_px = p_px;
        c_py = p_py;
        p_px = node[tmp].x;
        p_py = node[tmp].y;
        node[tmp].x = node[x].x - c_px;
        node[tmp].y = node[x].y - c_py;
        pre[x] = root;
        x = tmp;
    }
    return root;
}
void union_set(const int& root1,const int& root2,
        const int& x,const int& y)
{
    pre[root2] += pre[root1];
    pre[root1] = root2;
    node[root1].x = x;
    node[root1].y = y;
    return ;
}
void print()
{
    for(int i = 1; i <= n; ++i)
        printf("(%d,%d) ",node[i].x,node[i].y);
    printf("%\n");
}
int input()
{
    int root1,root2,x,y,i,j,a,b,len;
    char c;
    for(i = 0; i < m; ++i)
    {
        scanf("%d %d %d %c",&road[i].a,&road[i].b,&len,&c);
        x = y = 0;
        if(c=='N')
            y = len;
        else if(c=='S')
            y = -len;
        else if(c=='W')
            x = -len;
        else
            x =  len;
        road[i].x = x;
        road[i].y = y;
    }
    IN(k);
    for(i = 0; i < k; ++i)
        scanf("%d%d%d",&qus[i].a,&qus[i].b,&qus[i].ind);
    return 0;
}
int work()
{
    int i,j,tmp,root1,root2,ans,a,b,t,x,y;
    int pre = 0;
    sort(qus,qus+k,cmp);
    for(i = 0; i < k; ++i)
    {
        a = qus[i].a;
        b = qus[i].b;
        t = qus[i].ind;
        for(; pre < t; ++pre)
        {
            root1 = find_set(road[pre].a);
            root2 = find_set(road[pre].b);
            if(root1 == root2)
                continue;
            x = road[pre].x - node[road[pre].a].x + node[road[pre].b].x;
            y = road[pre].y - node[road[pre].a].y + node[road[pre].b].y;
            union_set(root1,root2,x,y);
        }
        root1 = find_set(a);
        root2 = find_set(b);
        ans = -1;
        if(root1 != root2)
            OUT(ans);
        else
        {
            root1 = ABS(node[a].x-node[b].x);
            root2 = ABS(node[a].y-node[b].y);
            ans = root1 + root2;
            OUT(ans);
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        input();
        work();
    }
    return 0;
}