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;
}