SPOJ GSS系列（数据结构维护技巧入门）

题目链接 GSS

$GSS1$

对于每个询问$l$, $r$，查询$a_{l}$, $a_{l+1}$, $a_{l+2}$, ..., $a_{r}$这个序列的最大字段和。

建立线段树，每个节点维护四个信息

$c$：当前区间的元素和

$lc$：当前区间左端点开始的最大子序列和

$rc$：当前区间右端点结束的最大子序列和

$ret$：当前区间的答案

于是我们建立线段树的时候预处理出每个节点的四个信息，查询的时候返回一个节点，这个节点的$ret$即为答案。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)

#define dec(i, a, b)	for (int i(a); i >= (b); --i)

#define lson		i << 1, L, mid

#define rson		i << 1 | 1, mid + 1, R

#define ls		i << 1

#define rs		i << 1 | 1

typedef long long LL;

const int N = 1e5 + 10;

struct node{

	int c, lc, rc, ret;

} t[N << 2];

int n, q, l, r;

void pushup(int i){

	t[i].c = t[ls].c + t[rs].c;

	t[i].lc = max(t[ls].lc, t[ls].c + t[rs].lc);

	t[i].rc = max(t[rs].rc, t[rs].c + t[ls].rc);

	t[i].ret = max(max(t[ls].ret, t[rs].ret), t[ls].rc + t[rs].lc);

}

void build(int i, int L, int R){

	if (L == R){

		scanf("%d", &t[i].ret);

		t[i].c = t[i].lc = t[i].rc = t[i].ret;

		return;

	}

	int mid = (L + R) >> 1;

	build(lson);

	build(rson);

	pushup(i);

}

node query(int i, int L, int R, int l, int r){

	if (l == L && R == r) return t[i];

	int mid = (L + R) >> 1;

	if (r <= mid) return query(lson, l, r);

	if (l > mid)  return query(rson, l, r);

	node ta = query(lson, l, mid);

	node tb = query(rson, mid + 1, r);

	node ans;

	ans.c = ta.c + tb.c;

	ans.lc = max(ta.lc, ta.c + tb.lc);

	ans.rc = max(tb.rc, tb.c + ta.rc);

	ans.ret = max(max(ta.ret, tb.ret), ta.rc + tb.lc);

	return ans;

}

int main(){

	while (~scanf("%d", &n)){

		build(1, 1, n);

		scanf("%d", &q);

		while (q--){

			scanf("%d%d", &l, &r);

			node ans = query(1, 1, n, l, r);

			printf("%d\n", ans.ret);

		}

	}

	return 0;

}

$GSS2$

$GSS1$的去重版

什么意思呢

询问的形式还是完全一样的，就是子序列的价值稍微变了一下。

$GSS1$中子序列的价值是该子序列的元素和，$GSS2$这里是该子序列出现过的元素的和

做法和$GSS1$完全不一样。

我们对询问离线。然后排序，按照右端点升序。

然后从左往右扫过去，每次加入$a_{i}$的时候， 先把$a_{i}$加进去。

接着我们对$last[a_{i}] + 1$到$i-1$这段区间都加上$a_{i}$，$last[x]$表示$x$上一次出现的位置，初值为$0$。

维护线段树的话我们需要四个信息。

$now$：当前这个时候区间的最大值

$mx$：历史区间的最大值

$h$：历史lazy标记（传给后代）的最大值

$add$：当前lazy标记（需传给后代）的值

然后查询一遍就可以了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)

#define dec(i, a, b)	for (int i(a); i >= (b); --i)

#define lson		i << 1, L, mid

#define rson		i << 1 | 1, mid + 1, R

#define ls		i << 1

#define rs		i << 1 | 1

#define MP		make_pair

#define fi		first

#define se		second

typedef long long LL;

const int N = 1e5 + 10;

const int d = 1e5;

int n, m, x, y;

int a[N], c[N << 1];

vector <pair <int, int> > v[N];

LL mx[N << 2], now[N << 2], add[N << 2], h[N << 2];

LL ans[N];

void pushup(int i){

	mx[i] = max(mx[ls], mx[rs]);

	now[i] = max(now[ls], now[rs]);

}

void pushdown(int i){

	h[ls] = max(h[ls], add[ls] + h[i]);

	h[rs] = max(h[rs], add[rs] + h[i]);

	add[ls] += add[i];

	add[rs] += add[i];

	mx[ls] = max(mx[ls], now[ls] + h[i]);

	mx[rs] = max(mx[rs], now[rs] + h[i]);

	now[ls] += add[i];

	now[rs] += add[i];

	add[i] = 0;

	h[i]   = -1e16;

}

void update(int i, int L, int R, int x, int val){

	if (L == R && L == x){

		mx[i] = now[i] = val;

		return;

	}

	int mid = (L + R) >> 1;

	pushdown(i);

	if (x <= mid) update(lson, x, val);

	else update(rson, x, val);

	pushup(i);

}

void Add(int i, int L, int R, int l, int r, int val){

	if (l <= L && R <= r){

		now[i] += val;

		mx[i] = max(mx[i], now[i]);

		add[i] += val;

		h[i] = max(h[i], add[i]);

		return ;

	}

	int mid = (L + R) >> 1;

	pushdown(i);

	if (l <= mid) Add(lson, l, r, val);

	if (r >  mid) Add(rson, l, r, val);

       	pushup(i);

}

LL query(int i, int L, int R, int l, int r){

	if (l <= L && R <= r) return mx[i];

	int mid = (L + R) >> 1;

	LL ret = -1e16;

	pushdown(i);

	if (l <= mid) ret = max(ret, query(lson, l, r));

	if (r > mid)  ret = max(ret, query(rson, l, r));

	pushup(i);

	return ret;

}

int main(){

	scanf("%d", &n);

	rep(i, 1, n) scanf("%d", a + i);

	scanf("%d", &m);

	rep(i, 1, ((n << 2) - 1)) mx[i] = h[i] = now[i] = -1e16;

	rep(i, 1, m){

		scanf("%d%d", &x, &y);

		v[y].push_back(MP(x, i));

	}

	rep(i, 1, n){

		update(1, 1, n, i, a[i]);

		if (c[a[i] + d] + 1 < i) Add(1, 1, n, c[a[i] + d] + 1, i - 1, a[i]);

		c[a[i] + d] = i;

		for (auto u : v[i]) ans[u.se] = query(1, 1, n, u.fi, i);

	}

	rep(i, 1, m) printf("%lld\n", max(0ll, ans[i]));

	return 0;

}

$GSS3$

$GSS1$加个单点更新，没了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)

#define dec(i, a, b)	for (int i(a); i >= (b); --i)

#define lson		i << 1, L, mid

#define rson		i << 1 | 1, mid + 1, R

#define ls		i << 1

#define rs		i << 1 | 1

typedef long long LL;

const int N = 1e5 + 10;

struct node{

	int c, lc, rc, ret;

} t[N << 2];

int n, q, l, r;

int op, x, y;

void pushup(int i){

	t[i].c = t[ls].c + t[rs].c;

	t[i].lc = max(t[ls].lc, t[ls].c + t[rs].lc);

	t[i].rc = max(t[rs].rc, t[rs].c + t[ls].rc);

	t[i].ret = max(max(t[ls].ret, t[rs].ret), t[ls].rc + t[rs].lc);

}

void build(int i, int L, int R){

	if (L == R){

		scanf("%d", &t[i].ret);

		t[i].c = t[i].lc = t[i].rc = t[i].ret;

		return;

	}

	int mid = (L + R) >> 1;

	build(lson);

	build(rson);

	pushup(i);

}

void update(int i, int L, int R, int x, int val){

	if (L == x && L == R){

		t[i].c = t[i].lc = t[i].rc = t[i].ret = val;

		return;

	}

	int mid = (L + R) >> 1;

	if (x <= mid) update(lson, x, val);

	else update(rson, x, val);

	pushup(i);

}

node query(int i, int L, int R, int l, int r){

	if (l == L && R == r) return t[i];

	int mid = (L + R) >> 1;

	if (r <= mid) return query(lson, l, r);

	if (l > mid)  return query(rson, l, r);

	node ta = query(lson, l, mid);

	node tb = query(rson, mid + 1, r);

	node ans;

	ans.c = ta.c + tb.c;

	ans.lc = max(ta.lc, ta.c + tb.lc);

	ans.rc = max(tb.rc, tb.c + ta.rc);

	ans.ret = max(max(ta.ret, tb.ret), ta.rc + tb.lc);

	return ans;

}

int main(){

	scanf("%d", &n);

	build(1, 1, n);

	scanf("%d", &q);

	while (q--){

		scanf("%d", &op);

		if (op == 1){

			scanf("%d%d", &l, &r);

			node ans = query(1, 1, n, l, r);

			printf("%d\n", ans.ret);

		}

		else{

			scanf("%d%d", &x, &y);

			update(1, 1, n, x, y);

		}

	}

	return 0;

}

$GSS4$

这题做法很暴力

对那些值已经全部是1的区间结点打上永久标记即可。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)

#define dec(i, a, b)	for (int i(a); i >= (b); --i)

#define ls		i << 1

#define rs		i << 1 | 1

#define lson		ls, L, mid

#define rson		rs, mid + 1, R

typedef long long LL;

const int N = 1e5 + 10;

LL s[N << 2];

int t[N << 2];

int n;

int q;

int ca = 0;

inline void pushup(int i){

	s[i] = s[ls] + s[rs];

	t[i] = t[ls] & t[rs];

}

void build(int i, int L, int R){

	if (L == R){

		scanf("%lld", s + i);

		t[i] = s[i] <= 1;

		return;

	}

	int mid = (L + R) >> 1;

	build(lson);

	build(rson);

	pushup(i);

}

void update(int i, int L, int R, int l, int r){

	if (t[i]) return;

	if (L == R){

		s[i] = sqrt(s[i]);

		t[i] = s[i] <= 1;

		return;

	}

	int mid = (L + R) >> 1;

	if (l <= mid) update(lson, l, r);

	if (r >  mid) update(rson, l, r);

	pushup(i);

}

LL query(int i, int L, int R, int l, int r){

	if (l <= L && R <= r) return s[i];

	int mid = (L + R) >> 1;

	LL ret = 0;

	if (l <= mid) ret += query(lson, l, r);

       	if (r > mid)  ret += query(rson, l, r);

	return ret;

}

int main(){

	while (~scanf("%d", &n)){

		printf("Case #%d:\n", ++ca);

		build(1, 1, n);

		scanf("%d", &q);

		while (q--){

			int op, x, y;

			scanf("%d%d%d", &op, &x, &y);

			if (x > y) swap(x, y);

			if (op == 1) printf("%lld\n", query(1, 1, n, x, y));

			else update(1, 1, n, x, y);

		}

		putchar(10);

	}

	return 0;

}

　　

$GSS5$

分类讨论，注意细节就可以了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)

#define dec(i, a, b)	for (int i(a); i >= (b); --i)

#define lson		i << 1, L, mid

#define rson		i << 1 | 1, mid + 1, R

#define ls		i << 1

#define rs		i << 1 | 1

const int N = 1e5 + 10;

struct node{

	int c, lc, rc, ret;

} t[N << 2];

int T;

int n, q, l, r;

int a[N], s[N];

int ans;

void pushup(int i){

	t[i].c = t[ls].c + t[rs].c;

	t[i].lc = max(t[ls].lc, t[ls].c + t[rs].lc);

	t[i].rc = max(t[rs].rc, t[rs].c + t[ls].rc);

	t[i].ret = max(max(t[ls].ret, t[rs].ret), t[ls].rc + t[rs].lc);

}

void build(int i, int L, int R){

	if (L == R){

		t[i].ret = a[L];

		t[i].c = t[i].lc = t[i].rc = t[i].ret;

		return;

	}

	int mid = (L + R) >> 1;

	build(lson);

	build(rson);

	pushup(i);

}

node query(int i, int L, int R, int l, int r){

	if (l == L && R == r) return t[i];

	int mid = (L + R) >> 1;

	if (r <= mid) return query(lson, l, r);

	if (l > mid)  return query(rson, l, r);

	node ta = query(lson, l, mid);

	node tb = query(rson, mid + 1, r);

	node ans;

	ans.c = ta.c + tb.c;

	ans.lc = max(ta.lc, ta.c + tb.lc);

	ans.rc = max(tb.rc, tb.c + ta.rc);

	ans.ret = max(max(ta.ret, tb.ret), ta.rc + tb.lc);

	return ans;

}

int main(){

	scanf("%d", &T);

	while (T--){

		scanf("%d", &n);

		rep(i, 1, n) scanf("%d", a + i);

		rep(i, 1, n) s[i] = s[i - 1] + a[i];

		build(1, 1, n);

		scanf("%d", &q);

		while (q--){

			int x1, y1, x2, y2;

			scanf("%d%d%d%d", &x1, &y1, &x2, &y2);

			ans = 0;

			if (x2 == y1){

				ans = query(1, 1, n, x1, y1).rc + query(1, 1, n, x2, y2).lc - a[y1];

				printf("%d\n", ans);

			}

			else if (y1 < x2){

				ans = s[x2 - 1] - s[y1];

				ans += query(1, 1, n, x1, y1).rc + query(1, 1, n, x2, y2).lc;

				printf("%d\n", ans);

			}

			else{

				ans = query(1, 1, n, x2, y1).ret;

				ans = max(ans, s[y1 - 1] - s[x2] + query(1, 1, n, x1, x2).rc + query(1, 1, n, y1, y2).lc);

				ans = max(ans, query(1, 1, n, x1, x2).rc + query(1, 1, n, x2, y1).lc - a[x2]);

				ans = max(ans, query(1, 1, n, x2, y1).rc + query(1, 1, n, y1, y2).lc - a[y1]);

				printf("%d\n", ans);

			}

		}

	}

	return 0;

}