POJ-1984 Navigation Nightmare

题目链接：Navigation Nightmare

题目大意：按照时间顺序给一些点的相对位置。有若干询问，每个询问给出两个点的编号以及时间，若在此时间内该两点相对位置确定则输出其曼哈顿距离，否则输出-1。

解题思路：很好的一道带权并查集问题，由于询问中的时间条件我纠结了很久，最后参考网上代码，按照询问的时间顺序来合并点并保存答案就行了。注意在更新相对位置的时候最好自己画一下图。

代码如下：

#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;}