P3527

Sol

整体二分板子，没啥好说。

讲一下一些细节:

1. 注意是区间加，最大总和可达 ${10}^{23}$，会爆 long long。
2. 注意复杂度，遍历出边次数最多为深度乘上边数，不会 TLE。

Code

#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define pf push_front
#define desktop "C:\\Users\\incra\\Desktop\\"
#define IOS ios :: sync_with_stdio (false),cin.tie (0),cout.tie (0)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair <int,int> PII;
const int dx[] = {1,0,-1,0},dy[] = {0,-1,0,1};
template <typename T1,typename T2> bool tomax (T1 &x,T2 y) {
	if (y > x) return x = y,true;
	return false;
}
template <typename T1,typename T2> bool tomin (T1 &x,T2 y) {
	if (y < x) return x = y,true;
	return false;
}
LL power (LL a,LL b,LL p) {
	LL ans = 1;
	while (b) {
		if (b & 1) ans = ans * a % p;
		a = a * a % p;
		b >>= 1;
	}
	return ans;
}
int fastio = (IOS,0);
#define endl '\n'
#define puts(s) cout << (s) << endl
const int N = 900010;
int n,m,k;
vector <int> g[N];
struct q_type {
    int k,id;
}qs[N],q1[N],q2[N];
struct BIT {
    LL c[N];
    int lowbit (int x) {
        return x & -x;
    }
    void modify (int x,LL d) {
        for (int i = x;i < N;i += lowbit (i)) c[i] += d;
    }
    LL query (int x) {
        LL ans = 0;
        for (int i = x;i;i -= lowbit (i)) ans += c[i];
        return ans;
    }
}c;
struct m_type {
    int l,r,d;
}ms[N];
int ans[N];
void solve (int l,int r,int lv,int rv) {
    if (l > r) return ;
    if (lv == rv) {
        for (int i = l;i <= r;i++) ans[qs[i].id] = lv;
        return ;
    }
    int mid = lv + rv >> 1;
    for (int i = lv;i <= mid;i++) c.modify (ms[i].l,ms[i].d),c.modify (ms[i].r + 1,-ms[i].d);
    int cnt1 = 0,cnt2 = 0;
    for (int i = l;i <= r;i++) {
        __int128 sum = 0;
        for (int j : g[qs[i].id]) sum += c.query (j) + c.query (j + m);
        if (sum >= qs[i].k) q1[++cnt1] = qs[i];
        else qs[i].k -= sum,q2[++cnt2] = qs[i];
    }
    for (int i = 1;i <= cnt1;i++) qs[l + i - 1] = q1[i];
    for (int i = 1;i <= cnt2;i++) qs[l + cnt1 + i - 1] = q2[i];
    for (int i = lv;i <= mid;i++) c.modify (ms[i].l,-ms[i].d),c.modify (ms[i].r + 1,ms[i].d);
    solve (l,l + cnt1 - 1,lv,mid),solve (l + cnt1,r,mid + 1,rv);
}
void mian () {
    cin >> n >> m;
    for (int i = 1;i <= m;i++) {
        int x;
        cin >> x;
        g[x].pb (i);
    }
    for (int i = 1;i <= n;i++) cin >> qs[i].k,qs[i].id = i;
    cin >> k;
    for (int i = 1;i <= k;i++) {
        int l,r,d;
        cin >> l >> r >> d;
        if (r < l) r += m;
        ms[i] = {l,r,d};
    }
    solve (1,n,1,k + 1);
    for (int i = 1;i <= n;i++) {
        if (ans[i] == k + 1) puts ("NIE");
        else cout << ans[i] << endl;
    }
}
signed main () {
	int T = 1;
	// cin >> T;
	while (T--) mian ();
	return 0;
}