2025CSP-S模拟赛31 比赛总结

2025CSP-S模拟赛31

T1	T2	T3	T4
0 RE	0 RE	0 RE	0 RE

总分：0；排名：17/17

比较神秘，没有加 freopen。挂分为 50，40，30，49。均为部分分。

T1 花海

T1 你这个关注到 \(n\times m \le 2\times 10^5\)，我们不妨假设 \(n\le m\)，则有 \(n\le \sqrt{2\times 10^5}\)。然后，我们考虑一个 \(n^2m\) 的算法，方法很多，随便口糊一个就好啦。

#include <bits/stdc++.h>
#define il inline

using namespace std;

const int bufsz = 1 << 20;
char ibuf[bufsz], *p1 = ibuf, *p2 = ibuf;
#define getchar() (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, bufsz, stdin), p1 == p2) ? EOF : *p1++)
il int read() {
	int x = 0; char ch = getchar(); bool t = 0;
	while (ch < '0' || ch > '9') {t ^= ch == '-'; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
	return t ? -x : x; 
}
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
int n, m;
vector<int> a[N], s[N];
il void init() {
	for (int i = 0; i <= n + 2; i++) {
		for (int j = 0; j <= m + 2; j++) {
			a[i].push_back(0);
			s[i].push_back(0);
		}
	}
}
il int getsum(int i, int j, int x, int y) {
	return i > x || j > y ? 0 : s[x][y] - s[x][j - 1] - s[i - 1][y] + s[i - 1][j - 1];
}
int main() {
	freopen("flower.in", "r", stdin);
	freopen("flower.out", "w", stdout);
	n = read(), m = read();
	int flag = 0;
	if (n > m) swap(n, m), flag = 1;
	init();
	if (!flag) {
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				a[i][j] = read();
			}
		}
	} else {
		for (int i = 1; i <= m; i++) {
			for (int j = 1; j <= n; j++) {
				a[j][i] = read();
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			s[i][j] = s[i - 1][j] + s[i][j - 1] + a[i][j] - s[i - 1][j - 1];
		}
	}
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		for (int x = i + 1; x <= n; x++) {
			int mx = -INF;
			for (int y = 1; y <= m; y++) {
				ans = max(ans, mx + getsum(i, 1, i, y) + getsum(x, 1, x, y) + getsum(i + 1, y, x - 1, y));
				mx = max(mx, getsum(i + 1, y, x - 1, y) - getsum(i, 1, i, y - 1) - getsum(x, 1, x, y - 1));
			}
		}
	}
	printf("%d\n", ans);
	
	return 0;
}

T2 划分

考虑 dp。

设 \(f_i\) 表示考虑到 \(i\) 的最小划分代价，令 \(s_i\) 为前缀和。显然有转移:

\[f_i=\min_{0\le j <i \and s_i-s_j\le m}\{f_j+\max_{k=j+1}^{i} a_k\} \]

然后令 \(p\) 表示第一个满足 \(s_j \geq s_i-m\) 的 \(j\)，那么式子便化为：

\[f_i=\min_{p \le j < i}\{f_j+\max_{k=j+1}^{i} a_k\} \]

考虑优化。首先，我们发现这个东西不是很好优化，然后考虑一个比较暴力的优化方式，线段树优化。考虑用线段树维护 \(f_j+\max_{k=j+1}^i a_k\)。其实这个不是很好维护。考试时维护这一坨维护了半天没写出来，赛后又写了会儿，前前后后写了 4h。其实有个更为简单好写的方式去维护，就是把这两项拆开维护，维护一个 \(f\) ，再维护一个 \(\max\)，再维护一个加和的最小值。这个就非常好写了。

#include <bits/stdc++.h>
#define il inline
#define int long long

using namespace std;

const int bufsz = 1 << 20;
char ibuf[bufsz], *p1 = ibuf, *p2 = ibuf;
#define getchar() (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, bufsz, stdin), p1 == p2) ? EOF : *p1++)
il int read() {
	int x = 0; char ch = getchar(); bool t = 0;
	while (ch < '0' || ch > '9') {t ^= ch == '-'; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
	return t ? -x : x;
} 
const int INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5 + 10;
int n, m, a[N], s[N];
int f[N];
int st[N], head, lst[N];
namespace Seg {
	struct node {
		int l, r, mn, lazy, mnf, mx;
	} tree[4 * N];
	#define lc p << 1
	#define rc p << 1 | 1
	il void pushup(int p) {
		tree[p].mn = min(tree[lc].mn, tree[rc].mn);
		tree[p].mnf = min(tree[lc].mnf, tree[rc].mnf);
	}
	il void build(int p, int l, int r) {
		tree[p].l = l, tree[p].r = r;
		if (l == r) return;
		int mid = l + r >> 1;
		build(lc, l, mid), build(rc, mid + 1, r);
	}
	il void pushdown(int p) {
		if (!tree[p].lazy) return;
		tree[lc].lazy = tree[p].lazy;
		tree[rc].lazy = tree[p].lazy;
		tree[lc].mx = tree[p].lazy;
		tree[rc].mx = tree[p].lazy;
		tree[lc].mn = tree[lc].mnf + tree[lc].mx;
		tree[rc].mn = tree[rc].mnf + tree[rc].mx;
		tree[p].lazy = 0;
	}
	il void update(int p, int l, int r, int v) {
		if (l > r) return;
		if (tree[p].l == l && tree[p].r == r) {
			tree[p].mx = v;
			tree[p].mn = tree[p].mnf + tree[p].mx;
			tree[p].lazy = v;
			return;
		}
		pushdown(p);
		int mid = tree[p].l + tree[p].r >> 1;
		if (r <= mid) update(lc, l, r, v);
		else if (l > mid) update(rc, l, r, v);
		else update(lc, l, mid, v), update(rc, mid + 1, r, v);
		pushup(p);
	}
	il void modify(int p, int x, int v) {
		if (tree[p].l == tree[p].r) {
			tree[p].mnf = v;
			tree[p].mn = tree[p].mnf + tree[p].mx;
			return;
		}
		pushdown(p);
		int mid = tree[p].l + tree[p].r >> 1;
		if (x <= mid) modify(lc, x, v);
		else modify(rc, x, v);
		pushup(p);
	}
	il int query(int p, int l, int r) {
		if (tree[p].l == l && tree[p].r == r) return tree[p].mn;
		pushdown(p);
		int mid = tree[p].l + tree[p].r >> 1;
		if (r <= mid) return query(lc, l, r);
		else if (l > mid) return query(rc, l, r);
		else return min(query(lc, l, mid), query(rc, mid + 1, r));
	}
}
signed main() {
	freopen("split.in", "r", stdin);
	freopen("split.out", "w", stdout);
	n = read(), m = read();
	for (int i = 1; i <= n; i++) {
		a[i] = read();
		s[i] = s[i - 1] + a[i];
	}
	for (int i = 1; i <= n; i++) {
		while (head && a[st[head]] < a[i]) head--;
		lst[i] = st[head];
		st[++head] = i;
	}
	Seg::build(1, 0, n);
	for (int i = 1; i <= n; i++) {
		int p = lower_bound(s, s + 1 + n, s[i] - m) - s;
		Seg::update(1, lst[i], i - 1, a[i]);
		f[i] = Seg::query(1, p, i - 1);
		Seg::modify(1, i, f[i]);
	}
	printf("%lld\n", f[n]);
	
	return 0;
}

T3 落子无悔

考虑把从上往下删反过来变成从下往上合并。

因为一个儿子对他的一个祖先的贡献是永远不会变的，所以考虑将这些点按一定顺序合并到父亲上，然后把合并后的点看作一个新的点，并在合并的过程中统计贡献。

考虑合并的顺序。对于两个点 \(a,b\) 来讲，令 \(s0,s1\) 分别表示 0 和 1 的数量。那么，如果 \(a\) 排在 \(b\) 前更优，则有 \(s1_a\times s0_b \le s1_b \times s0_a\)。

然后把所有点排一下统计贡献即可。根节点要特判。

#include <bits/stdc++.h>
#define il inline
#define int long long

using namespace std;

const int bufsz = 1 << 20;
char ibuf[bufsz], *p1 = ibuf, *p2 = ibuf;
#define getchar() (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, bufsz, stdin), p1 == p2) ? EOF : *p1++)
il int read() {
	int x = 0; char ch = getchar(); bool t = 0;
	while (ch < '0' || ch > '9') {t ^= ch == '-'; ch = getchar();}
	while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar();}
	return t ? -x : x; 
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 10;
int n, f[N], a[N];
int son[N];
int s0[N], s1[N];
struct node {
	int id, s0, s1;
	bool operator < (const node & cmp) const {
		return s1 * cmp.s0 != cmp.s1 * s0 ? s1 * cmp.s0 < cmp.s1 * s0 : id < cmp.id;
	}
};
set<node> st;
int fa[N];
il int getfa(int x) {
	return x == fa[x] ? x : fa[x] = getfa(fa[x]);
}
signed main() {
	freopen("board.in", "r", stdin);
	freopen("board.out", "w", stdout);
	n = read();
	for (int i = 2; i <= n; i++) {
		f[i] = read();
	}
	for (int i = 1; i <= n; i++) {
		a[i] = read();
		s0[i] = (a[i] == 0);
		s1[i] = (a[i] == 1);
	}
	for (int i = 2; i <= n; i++) {
		st.insert((node){i, s0[i], s1[i]});
	}
	for (int i = 1; i <= n; i++) {
		fa[i] = i;
	}	
	int ans = 0;
	while (!st.empty()) {
		node t = *(st.begin());
		st.erase((node){t.id, t.s0, t.s1});
		int x = getfa(f[t.id]);
		if (x != 1) st.erase((node){x, s0[x], s1[x]});
		ans += s1[x] * t.s0;
		s0[x] += t.s0, s1[x] += t.s1;
		fa[t.id] = x;
		if (x != 1) st.insert((node){x, s0[x], s1[x]});
	}
	printf("%lld\n", ans);
	
	return 0;
}

T4 体测

posted @ 2025-08-09 14:15 Zctf1088 阅读(25) 评论(0) 收藏举报

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