POJ-1984 Navigation Nightmare

題目鏈接：Navigation Nightmare

題目大意：按照時間順序給一些點的相對位置。有若干詢問，每個詢問給出兩個點的編號以及時間，若在此時間內該兩點相對位置確定則輸出其曼哈頓距離，否則輸出-1。

解題思路：很好的一道帶權並查集問題，由於詢問中的時間條件我糾結了很久，最後參考網上代碼，按照詢問的時間順序來合併點並保存答案就行了。注意在更新相對位置的時候最好自己畫一下圖。

代碼如下：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int inf  = 0x3f3f3f3f;
const int maxm = 1e4 + 15;
const int maxn = 4e4 + 15;

struct Node{
    int x, y, id, t;    
    bool operator < (const Node& rhs) const {
        return t < rhs.t;   
    }
}node[maxm];

int dx[maxn], dy[maxn], rx[maxn], ry[maxn], x[maxn], y[maxn];
int par[maxn], ans[maxm];

int find(int x) {
    if(par[x] == x) return x;   
    int res = find(par[x]);
    int fa = par[x];
    // root -> fa + fa -> x ==> root -> x 
    rx[x] = rx[fa] + rx[x];
    ry[x] = ry[fa] + ry[x];
    return par[x] = res;
}

inline int read() {
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c)) {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

int main(){
#ifdef NEKO
    freopen("Nya.txt", "r", stdin);
#endif
    int n, m, k; cin >> n >> m;
    for(int i = 1; i <= n; i++) {
        par[i] = i; 
        // relative position
        rx[i] = ry[i] = 0;
    }

    for(int i = 1; i <= m; i++) {
        int d; char c;
        x[i] = read(); y[i] = read(); d = read();
        scanf("%c", &c);
        switch(c) {
            case 'W': dx[i] = -d; dy[i] =  0; break;        
            case 'S': dx[i] =  0; dy[i] = -d; break;        
            case 'E': dx[i] =  d; dy[i] =  0; break;        
            case 'N': dx[i] =  0; dy[i] =  d; break;        
        }
    }
    k = read();
    for(int i = 1; i <= k; i++) {
        node[i].x = read(); node[i].y = read();
        node[i].t = read(); node[i].id = i;     
    }
    sort(node + 1, node + 1 + k);
    int j = 1;
    for(int i = 1; i <= k; i++) {
        for(; j <= node[i].t; j++) {
            int l = find(x[j]), r = find(y[j]); 
            if(l != r) {
                par[l] = r;
                // root -> l <== root -> y + y -> x + x -> l
                rx[l] = rx[y[j]] + dx[j] - rx[x[j]];
                ry[l] = ry[y[j]] + dy[j] - ry[x[j]];
            }   
        }
        if(find(node[i].x) != find(node[i].y))
            ans[node[i].id] = -1;
        else
            ans[node[i].id] = abs(rx[node[i].x] - rx[node[i].y])
            + abs(ry[node[i].x] - ry[node[i].y]);
    }
    for(int i = 1; i <= k; i++)
        printf("%d\n", ans[i]);
    return 0;
}