HDU 4879 ZCC loves march 【map+set+并查集】

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题目链接

题意：在矩阵中有n个点，现有两种操作。Q x表示把x这个点所在行以及所在列的所有点都并到这个点上，并输出所有点移动距离的平方和。另外就是把某个点向某个方向移动若干位。

用两个$map<ll,set<ll>>$ 分别存每行中点的集合和每列中点的集合，再用并查集的根节点存每个点所在的集合。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn = 200000 + 10;
const ll MOD = 1e9 + 7;

ll par[maxn];
/*
ll find(ll x) {
    if (x == par[x])
        return x;
    else
        return par[x] = find(par[x]);
}
*/

ll find(ll x)
{
    if (par[x] != x)
        par[x] = find(par[x]);
    return par[x];
}

struct Node {
    ll x, y;
    ll num;
    Node() {}
    Node(ll _x, ll _y, ll _num): x(_x), y(_y), num(_num) {}
} node[maxn];

ll n, m;
ll ans = 0;
ll cnt = 0;
ll T;
char s[5];
map < ll, set<ll> > sX;
map < ll, set<ll> > sY;

ll rt, tx, ty, tnum, tmp, a, b, x, y;

int main() {
    scanf("%lld %lld", &n, &m);
    sX.clear();
    sY.clear();
    for (ll i = 1; i <= n; i++) {
        scanf("%lld %lld", &node[i].x, &node[i].y);
        sX[node[i].x].insert(i);
        sY[node[i].y].insert(i);
        // printf("%lld %lld %lld\n", i, node[i].x, node[i].y);
        par[i] = i;
        node[i].num = 1;
    }
    scanf("%lld", &T);
    ans = 0;
    cnt = n + 1;
    while (T--) {
        scanf("%s", s);
        if (s[0] == 'Q') {
            scanf("%lld", &rt);
            rt ^= ans;
            // printf("== %lld\n", rt);
            rt = find(rt);
            // printf("== %lld\n", rt);
            tx = node[rt].x;
            ty = node[rt].y;
            tnum = 0;
            ans = 0;
            set <ll>::iterator it;
            for (it = sX[tx].begin(); it != sX[tx].end(); it++) {
                // printf("%lld\n", *it);
                tmp = abs(node[*it].y - ty);
                tmp %= MOD;
                // printf("it = %lld  tmp = %lld  num = %lld\n", *it, tmp, node[*it].num);
                ans = (ans + tmp * tmp % MOD * node[*it].num % MOD + MOD) % MOD;
                tnum += node[*it].num;
                par[*it] = cnt;
                sY[node[*it].y].erase(*it);
            }
            // printf("============================\n");
            for (it = sY[ty].begin(); it != sY[ty].end(); it++) {
                // printf("%lld\n", *it);
                tmp = abs(node[*it].x - tx);
                tmp %= MOD;
                // printf("it = %lld  tmp = %lld  num = %lld\n", *it, tmp, node[*it].num);
                ans = (ans + tmp * tmp % MOD * node[*it].num % MOD + MOD) % MOD;
                tnum += node[*it].num;
                par[*it] = cnt;
                sX[node[*it].x].erase(*it);
            }
            par[cnt] = cnt;
            node[cnt] = Node(tx, ty, tnum);
            sX[tx].clear();
            sY[ty].clear();
            sX[tx].insert(cnt);
            sY[ty].insert(cnt);
            cnt++;
            printf("%lld\n", ans);
        }
        else {
            scanf("%lld %lld", &a, &b);
            a ^= ans;
            // printf("== %lld\n",a);
            rt = find(a);
            // printf("=== %lld\n",rt);
            // printf("=== %lld\n",find(4));
            x = node[rt].x;
            y = node[rt].y;
            sX[x].erase(a);
            sY[y].erase(a);
            node[rt].num--;
            if (node[rt].num == 0) {
                sX[x].erase(rt);
                sY[y].erase(rt);
            }
            if (s[0] == 'L') {
                tx = x;
                ty = y - b;
                // printf("tx = %lld  ty = %lld\n", tx, ty);
                node[a] = Node(tx, ty, 1);
                sX[tx].insert(a);
                sY[ty].insert(a);
                par[a] = a;
            }
            if (s[0] == 'R') {
                tx = x;
                ty = y + b;
                node[a] = Node(tx, ty, 1);
                sX[tx].insert(a);
                sY[ty].insert(a);
                par[a] = a;
            }
            if (s[0] == 'U') {
                tx = x - b;
                ty = y;
                node[a] = Node(tx, ty, 1);
                sX[tx].insert(a);
                sY[ty].insert(a);
                par[a] = a;
            }
            if (s[0] == 'D') {
                tx = x + b;
                ty = y;
                node[a] = Node(tx, ty, 1);
                sX[tx].insert(a);
                sY[ty].insert(a);
                par[a] = a;
            }
        }
    }
}