BZOJ 2527 [Poi2011]Meteors (整体二分+树状数组)

整体二分板题,没啥好讲的…注意是个环…还有所有贡献会爆longlong,那么只要在加之前判断一下有没有达到需要的值就行了…

CODE

#include <set>
#include <queue>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
char cb[1<<15],*cs=cb,*ct=cb;
#define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
template<class T>inline void read(T &res) {
    char ch; int flg = 1; for(;!isdigit(ch=getchar());)if(ch=='-')flg=-flg;
    for(res=ch-'0';isdigit(ch=getchar());res=res*10+ch-'0'); res*=flg;
}
const int MAXN = 300005;
vector<int>vec[MAXN];
int n, m, k, O[MAXN], L[MAXN], R[MAXN], A[MAXN], ans[MAXN];
struct query { int p, id; }q[MAXN], tmp[MAXN];
LL T[MAXN], val[MAXN];
inline void upd(int x, int val) {
	while(x <= m) T[x] += val, x += x&-x;
}
inline LL qsum(int x) { LL re = 0;
	while(x) re += T[x], x -= x&-x;
	return re;
}
inline void calc(int ql, int qr, int l, int r) {
	for(int i = ql; i <= qr; ++i) val[i] = 0;
	for(int i = l; i <= r; ++i) {
		upd(L[i], A[i]), upd(R[i]+1, -A[i]);
		if(L[i] > R[i]) upd(1, A[i]);
	}
	for(int i = ql; i <= qr; ++i)
		for(int j = 0, siz = vec[q[i].id].size(); j < siz; ++j) {
			val[i] += qsum(vec[q[i].id][j]);
			if(val[i] >= q[i].p) break;
		}
	for(int i = l; i <= r; ++i) {
		upd(L[i], -A[i]), upd(R[i]+1, A[i]);
		if(L[i] > R[i]) upd(1, -A[i]);
	}
}

void solve(int ql, int qr, int vl, int vr) {
	if(qr < ql) return;
	if(vl == vr) { for(int i = ql; i <= qr; ++i) ans[q[i].id] = vl; return; }
	int vmid = (vl + vr) >> 1;
	calc(ql, qr, vl, vmid);
	int st = ql, ed = qr;
	for(int i = ql; i <= qr; ++i)
		if(q[i].p <= val[i]) tmp[st++] = q[i];
		else q[i].p -= val[i], tmp[ed--] = q[i];
	for(int i = ql; i <= qr; ++i) q[i] = tmp[i];
	solve(ql, st-1, vl, vmid);
	solve(ed+1, qr, vmid+1, vr);
}

int main () {
	read(n), read(m);
	for(int i = 1, x; i <= m; ++i) read(x), vec[x].push_back(i);
	for(int i = 1; i <= n; ++i) read(q[i].p), q[i].id = i;
	read(k);
	for(int i = 1; i <= k; ++i)
		read(L[i]), read(R[i]), read(A[i]);
	solve(1, n, 1, k+1);
	for(int i = 1; i <= n; ++i)
		if(ans[i] <= k) printf("%d
", ans[i]);
		else puts("NIE");
}