55分代码

#define  _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include<map>
#include <string>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <iomanip>
#include<stack>
#include<queue>
#include<set>
#include<unordered_map>
#include <climits>
#include <math.h>
//众所周知，set能有序地维护同一类型的元素，但相同的元素只能出现一次。
using namespace std;
typedef long long LL;
const int N = 1e3,mod=1e9+7;
int dex[] = { 1,1,0,-1,-1,-1,0,1 };
int dey[] = { 0,1,1,1,0,-1,-1,-1 };
int n, m, p, q;
int ui, vi, ti;
unordered_map<LL, int>dict;
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> p >> q;
    for (int i = 0; i < p; i++) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        dict[(LL)y * n + x] = i + 1;
    }
    while (q--)
    {
        cin >> ui >> vi >> ti;
        ui--, vi--;
        int dist = INT_MAX;
        int mindist = min(min(ui , n - ui - 1), min(vi, m - vi - 1));
        if(n<10000)
        for (int i = 0; i < 8; i++) {
            for (int j = 1;; j++) {
                int x = ui + j * dex[i];
                int y = vi + j * dey[i];
                if (x >= n || x < 0 || y < 0 || y >= m) break;
                if (dict[(LL)n * y + x] != 0) {
                    dist = min(dist, j);
                    break;
                }
            }
        }
        else {
            for (auto &item : dict) {
                if (item.second == 0) continue;
                int y = item.first / n;
                int x = item.first % n;
                if (y == vi && x!=ui) dist = min(dist, abs(x - ui));
                else if (x == ui && y!=vi) dist = min(dist, abs(y - vi));
                else if(abs(x-ui)!=0 && abs(x-ui)==abs(y-vi)) {
                    dist = min(dist, abs(x - ui));
                }
            }
        }
        //cout << dist << endl;
        if (dist > mindist) continue;
        vector<int> td;
        for (int i = 0; i < 8; i++) {
            td.push_back(dict[n * ((LL)dey[i] * dist + vi) + (LL)dist * dex[i] + ui]);
            dict.erase(n * ((LL)dey[i] * dist + vi) + (LL)dist * dex[i] + ui);
        }
        for (int i = 0; i < 8; i++) {
            int next = (i + ti) % 8;
            int tx = ui + dex[next] * dist;
            int ty = vi + dey[next] * dist;
            LL keyy = (LL)ty * n + tx;
            if (td[i] != 0) {
                dict[keyy] = td[i];
            }
        }
    }
    LL ans = 0;
    for (auto& item : dict) {
        if (item.second != 0) {
            int y = item.first / n;
            int x = item.first % n;
            //cout <<"id:" << item.second << " x:" << x + 1 << " y:" << y + 1 << endl;
            ans =ans^( (LL)item.second * (x + 1) + y + 1);
        }
    }
    cout << ans << endl;
    return 0;
}