NOIP做题练习（day5）

A - 中位数图

题面

题解

先找出题意中的\(b\)所在的位置。

再以这个位置为中心，向右\(for\)一遍有多少个大于/小于该数的数

大于就\(++cs\) 小于就\(--cs\)。

因为这个数是中位数，所以大于它的数小于它的数的数量是一样的。

然后每次以\(cs\)为下标的数\(++\)。

因为\(cs\)可能小于零，

所以需要将\(cs\)加上\(100000\)后再\(++\)。

统计答案时直接加上下标为\(0-cs+100000\)的数的值即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#define int long long
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int n, b, a[100003], p[100003], c, s[100003], cs, cj, sum, ans, pp[200003];

signed main()
{
	//File("A");
	n = gi(), b = gi();
	for (itn i = 1; i <= n; i+=1)
	{
		a[i] = gi();
		if (a[i] == b) c = i;
	}
	for (int i = c; i <= n; i+=1)
	{
		if (a[i] > b) ++cs;
		if (a[i] < b) --cs;
		++pp[cs + 100000];
	}
	for (int i = c; i >= 1; i-=1)
	{
		if (a[i] > b) ++cj;
		if (a[i] < b) --cj;
		ans = ans + pp[0 - cj + 100000];
	}
	printf("%lld\n", ans);
	return 0;
}

B - 树

题面

题解

直接暴力\(dfs\)即可\(AC\)。

数据很水啊……

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int n, m, s, a[100003], fa[100003], ans;
int tot, head[300003], nxt[300003], ver[300003];

inline void add(int u, int v)
{
	ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

void dfs(int u, int fa, int w)
{
	if (w > s) return;
	if (w == s) {++ans; return;}
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == fa) continue;
		dfs(v, u, w + a[v]);
	}
}

int main()
{
	//File("B");
	n = gi(), s = gi();
	for (int i = 1; i <= n; i+=1) a[i] = gi();
	for (int i = 1; i < n; i+=1)
	{
		int u = gi(), v = gI();
		add(u, v);
		fa[v] = u;
	}
	for (int i = 1; i <= n; i+=1)
	{
		dfs(i, fa[i], a[i]);
	}
	printf("%d\n", ans);
	return 0;
}

C - 松鼠的新家

题面

题解

倍增\(LCA+\)树上差分一遍过。

注意要判断两个房间是不是存在祖先关系。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <vector>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

itn n, a[300003], dep[300003], w[300003], fa[300003][23], sum, ans, tot, head[1200003], nxt[1200003], ver[1200003];
int f[300003], logn;

inline void add(int u, int v)
{
	ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

void dfs(int u, int baba, int step)
{
	dep[u] = step; fa[u][0] = baba;
	for (int i = 1; i <= logn; i+=1)
	{
		fa[u][i] = fa[fa[u][i - 1]][i - 1];
	}
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == baba) continue;
		dfs(v, u, step + 1);
	}
}

inline void getlca(int u, int v)
{
	if (dep[u] < dep[v]) swap(u, v);
	int x = u, y = v;
	while (dep[u] > dep[v])
	{
		for (int i = logn; i >= 0; i-=1)
		{
			if (dep[fa[u][i]] > dep[v]) u = fa[u][i];
		}
		u = fa[u][0];
	}
	if (u == v)
	{
		++w[x], --w[fa[u][0]]; return;
	}
	for (itn i = logn; i >= 0; i-=1)
	{
		if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
	}
	u = fa[u][0];
	++w[x], ++w[y], --w[u], --w[fa[u][0]];
}

void getans(int u)
{
	f[u] = w[u];
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == fa[u][0]) continue;
		getans(v);
		f[u] = f[u] + f[v];
	}
}

int main()
{
	//File("C");
	n = gi();
	while ((1 << logn) < n) ++logn;
	for (int i = 1; i <= n; i+=1)
	{
		a[i] = gi();
	}
	for (int i = 1; i < n; i+=1)
	{
		int u = gi(), v = gi();
		add(u, v); add(v, u);
	}
	add(0, a[1]);
	dfs(0, 0, 1);
	for (int i = 1; i < n; i+=1)
	{
		getlca(a[i], a[i + 1]);
	}
	getans(a[1]);
	for (int i = 1; i <= n; i+=1)
	{
		printf("%d\n", (i == a[1]) ? (f[i]) : (f[i] - 1));
	}
	return 0;
}

总结

为什么每次时间内\(0AC\)，时间结束之后再切题呢？

思维要更活跃一点啊。