[luogu P3527] [POI2011]MET-Meteors

P3527 [POI2011]MET-Meteors

题解

对于每个国家进行二分，判断
发现时间复杂度十分垃圾$O(nmlogm)$,比暴力还慢
考虑整体二分
对所有国家的时间一起二分，mid时间
计算每个国家收集的陨石数量，吧能够完成的国家扔一遍，不能完成的扔另一边
左边右边再继续二分下去
每一层都是$O(nlogn)logk层$,所以时间复杂度就是两个log

// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define lowbit(x) (x & -x)
#define int long long
#define N 1000005
using namespace std;
struct A{
	int nd, id;
}b[N], p1[N], p2[N];
struct Q{
	int l, r, x;
}q[N];
int tree[N];
vector<int> g[N];
void update(int x, int y){ //printf("%lld _ %lld\n", x, y);
	for(;x < N; x += lowbit(x)) tree[x] += y;
}
int query(int x){
	int ret = 0;
	for(; x; x -= lowbit(x)) ret += tree[x];
	return ret;
}
int ans[N], a[N], n, m, tot;
void ef(int l, int r, int x, int y){
	if(x > y) return;
	if(l == r){
		for(int i = x; i <= y; i ++) ans[b[i].id] = l;
		return;
	}
	int len1 = 0 , len2 = 0;
	int mid = (l + r) >> 1;
	for(int i = l; i <= mid; i ++) update(q[i].l, q[i].x), update(q[i].r + 1, - q[i].x);
	for(int i = x; i <= y; i ++){
		int tmp = 0;
		for(int j = 0; j < g[b[i].id].size(); j ++){
			tmp += query(g[b[i].id][j]) + query(g[b[i].id][j] + m);
			if(tmp >= b[i].nd) break;
		}
		if(tmp >= b[i].nd) p1[++ len1] = b[i];
		else p2[++ len2] = b[i], p2[len2].nd -= tmp;
	}
//	printf("* %lld %lld %lld %lld\n", l, r, x, y);
//	for(int i = 1; i <= 20; i ++) printf("%lld ", query(i)); printf(" &\n");
//	for(int i = 1; i <= len1; i ++) printf("%lld ", p1[i].id); printf("\n");
//	for(int i = 1; i <= len2; i ++) printf("%lld ", p2[i].id); printf("\n\n");
	for(int i = l; i <= mid; i ++) update(q[i].l, - q[i].x), update(q[i].r + 1, q[i].x);
	for(int i = 1; i <= len1; i ++) b[i + x - 1] = p1[i];
	for(int i = 1; i <= len2; i ++) b[i + x + len1 - 1] = p2[i];
	ef(l, mid, x, x + len1 -1), ef(mid + 1, r, x + len1, y);
}
signed main(){
	scanf("%lld%lld", &n, &m);
	for(int i = 1; i <= m; i ++) scanf("%lld", &a[i]), g[a[i]].push_back(i);
	for(int i = 1; i <= n; i ++) scanf("%lld", &b[i].nd), b[i].id = i;
	scanf("%lld", &tot);
	for(int i = 1; i <= tot; i ++){
		scanf("%lld%lld%lld", &q[i].l, &q[i].r, &q[i].x);
		if(q[i].l > q[i].r) q[i].r += m;
	}
	q[++ tot].l = 1, q[tot].r = n * 2, q[tot].x = 1e16; 
	ef(1, tot, 1, n);
	for(int i = 1; i <= n; i ++) if(ans[i] != tot) printf("%lld\n", ans[i]); else printf("NIE\n");	
	return 0;
}