题解

之前我发啥电用广搜做啊，直接记录每一个敌军被秒的时间不就行了

#include <iostream>
#include <cstdio>
#include <queue>
#include <memory.h>
using namespace std;
struct node{
    int x, y, t;
}a[5005], b[5005];
int n, m, k, x, y, t;
int get(int x, int y){
    int mn = 0x3f3f3f3f;
    for(int i = 1; i <= k; i++){
        mn = min(mn, abs(b[i].x - x) + abs(b[i].y - y) + b[i].t);
    }
    return mn + t;
}
    
int main(){
    cin >> n >> m >> k;
    for(int i = 1; i <= m; i++){
        cin >> a[i].x >> a[i].y;
    }
    for(int i = 1; i <= k; i++){
        cin >> b[i].x >> b[i].y >> b[i].t;
    }
    int mx = 0;
    for(int i = 1; i <= m; i++){
        mx = max(mx, get(a[i].x, a[i].y));
    }
	cout << mx;
    
    return 0;
}