$读入数量庞大，必须用scanf读入$

$cin已经TLE了$

#include <iostream>
#include <queue>
#include <vector>
#include <cstring>

#define x first
#define y second

using namespace std;

const int M = 1e6+10, N = 1010;

typedef pair<int, int> PII;
typedef pair<PII, int> PIII;
typedef long long LL;

PIII s[M];  //存放客户信息
PII h[M];     //存放分店位置信息
int g[N][N];
int dist[N][N];
int n, m, k, d;
LL ans;             //答案记得开long long

void bfs()
{
    int dx[] = {1,0,-1,0}, dy[] = {0,-1,0,1};
    queue<PII> q;
    for(auto t : h){   //将所有分店入队
        q.push(t);
        dist[t.x][t.y] = 0;
    }

    while(q.size())
    {
        auto t = q.front();
        q.pop();

        for(int i = 0; i < 4; i ++)
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if(a < 1 || a > n || b < 1 || b > n) continue;
            if(g[a][b] || dist[a][b] != -1) continue;

            dist[a][b] = dist[t.x][t.y] + 1;
            q.push({a, b});
        }
    }

    for(auto t : s){     //最后将到达每个客户的最短距离乘上订单数，再求和即为答案
        ans += dist[t.x.x][t.x.y] * t.y;
    }
}

int main()
{
    memset(dist, -1, sizeof dist);

    scanf("%d%d%d%d",&n, &m, &k, &d);
    for(int i = 0; i < m; i ++){   //读入分店位置
        int a, b;
        scanf("%d%d",&a, &b);
        h[i] = {a, b};
    }

    for(int i = 0; i < k; i ++){   //读入客户信息
        int a, b, v;
        scanf("%d%d%d",&a, &b, &v);
        s[i] = {{a, b}, v};
    }

    for(int i = 0; i < d; i ++){    //读入不能走的坐标
        int a, b;
        scanf("%d%d",&a, &b);
        g[a][b] = 1;     // 1 表示不能走
    }

    bfs();

    cout << ans << endl;

    return 0;
}