省选模拟（12.08） T1 逐梦逐梦逐梦

逐梦逐梦逐梦

题目背景：

12.08 省选模拟T1

分析：主席树

 

这个题，除了细节太多，代码太难写，实现起来比较烦以外······其他还是很简单的。需要支持区间前k大求和，以及区间查找第k个偶数或者奇数，那么我们需要维护的就是当前权值区间的所有数的个数，奇数的个数，偶数的个数，以及所有数的和，因为是区间上的查询，所以外面套主席树就可以了，对于一次询问，先求当前区间的前k大之和，以及前k个中有多少个奇数，多少个偶数，如果和为偶数则就是答案，否则需要查找下一个奇数，以及下一个偶数，减去和中的最后一个奇数，然后加上下一个偶数，或者减去和中的最后一个偶数，加上下一个奇数，两者中的较大值，即为答案，复杂度大常数O(nlogn)

  

Source：

/*
	created by scarlyw
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

inline char read() {
	static const int IN_LEN = 1024 * 1024;
	static char buf[IN_LEN], *s, *t;
	if (s == t) {
		t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
		if (s == t) return -1;
	}
	return *s++;
}

///*
template
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = read(), iosig = false; !isdigit(c); c = read()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;	
	}
	for (x = 0; isdigit(c); c = read()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {
	if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;
	*oh++ = c;
}


template
inline void W(T x) {
	static int buf[30], cnt;
	if (x == 0) write_char('0');
	else {
		if (x < 0) write_char('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) write_char(buf[cnt--]);
	}
}

inline void flush() {
	fwrite(obuf, 1, oh - obuf, stdout);
}

/*
template
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar())
		if (c == '-') iosig = true;	
	for (x = 0; isdigit(c); c = getchar()) 
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}
//*/

const int MAXN = 300000 + 10;

struct node {
	int left, right;
	long long tot_s;
	int tot_c, cnt[2];
} tree[MAXN * 20];

struct data {
	int num, ori;
	inline bool operator < (const data &a) const {
		return num > a.num;
	}
} a[MAXN];

int cnt;
int rk[MAXN], root[MAXN], b[MAXN];

inline void insert(int &cur, int l, int r, int x) {
	tree[++cnt] = tree[cur], cur = cnt, tree[cur].tot_s += (long long)rk[x];
	tree[cur].tot_c++, tree[cur].cnt[rk[x] & 1]++;
	if (l == r) return ;
	int mid = l + r >> 1;
	if (x <= mid) insert(tree[cur].left, l, mid, x);
	else insert(tree[cur].right, mid + 1, r, x);
}

int c[2];
long long ans;
inline void query_k_sum(int l, int r, int ql, int qr, int k) {
	if (tree[qr].tot_c - tree[ql].tot_c == k) {
		c[0] += tree[qr].cnt[0] - tree[ql].cnt[0];
		c[1] += tree[qr].cnt[1] - tree[ql].cnt[1];
		ans += tree[qr].tot_s - tree[ql].tot_s;
		return ;
	}
	if (k == 0) return ;
	int mid = l + r >> 1, ll = tree[ql].left, rl = tree[qr].left;
	int temp = tree[rl].tot_c - tree[ll].tot_c;
	if (temp <= k) {
		c[0] += tree[rl].cnt[0] - tree[ll].cnt[0];
		c[1] += tree[rl].cnt[1] - tree[ll].cnt[1];
		ans += tree[rl].tot_s - tree[ll].tot_s;
		query_k_sum(mid + 1, r, tree[ql].right, tree[qr].right, k - temp);
	} else query_k_sum(l, mid, tree[ql].left, tree[qr].left, k);
}

inline int query_kth(int l, int r, int ql, int qr, int k, int type) {
	if (l == r) return rk[l];
	int mid = l + r >> 1, ll = tree[ql].left, rl = tree[qr].left;
	int temp = tree[rl].cnt[type] - tree[ll].cnt[type];
	if (temp >= k) {
		return query_kth(l, mid, tree[ql].left, tree[qr].left, k, type);
	} else return query_kth(mid + 1, r, tree[ql].right, 
		tree[qr].right, k - temp, type);
}

int n, q, l, r, k;
inline void solve_tree() {
	R(n);
	for (int i = 1; i <= n; ++i) R(a[i].num), a[i].ori = i;
	std::sort(a + 1, a + n + 1);
	for (int i = 1; i <= n; ++i) b[a[i].ori] = i, rk[i] = a[i].num;
	for (int i = 1; i <= n; ++i) 
		root[i] = root[i - 1], insert(root[i], 1, n, b[i]);
}

inline void solve_query() {
	R(q);
	while (q--) {
		R(l), R(r), R(k), ans = c[0] = c[1] = 0;
		if (k > r - l + 1 || (k & 1)) {
			W(-1), write_char('\n');
			continue ;
		}
		query_k_sum(1, n, root[l - 1], root[r], k);
		if ((c[0] & 1) || (c[1] & 1)) {
			int sum[2];
			long long ret = -1;
			sum[0] = tree[root[r]].cnt[0] - tree[root[l - 1]].cnt[0];
			sum[1] = tree[root[r]].cnt[1] - tree[root[l - 1]].cnt[1];
			if (sum[0] != c[0]) {
				int x1 = query_kth(1, n, root[l - 1], root[r], c[1], 1);
				int x2 = query_kth(1, n, root[l - 1], root[r], c[0] + 1, 0);
				ret = std::max(ret, ans - x1 + x2);
			}
			if (sum[1] != c[1]) {
				int x1 = query_kth(1, n, root[l - 1], root[r], c[0], 0);
				int x2 = query_kth(1, n, root[l - 1], root[r], c[1] + 1, 1);
				ret = std::max(ret, ans - x1 + x2);
			}
			W(ret), write_char('\n');
		} else W(ans), write_char('\n');
	}
}

int main() {
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);
	solve_tree();
	solve_query();
	flush();
	return 0;
}