洛谷 P5021 [NOIP2018]赛道重建

洛谷 P5021 [NOIP2018]赛道重建

传送门

思路

思路就是常规的思路，所以就不说了……我就是来记录一下我的\(AC\)之路的，真的是太爽了

没错……我也是一个个打的部分分，最后终于AC的，至于为什么中间又会有\(35\)、\(25\)、\(0\)这样的分数……纯粹是因为我犯了zz错误……

代码

1、\(b_i = a_i + 1\) 链的情况

#include <bits/stdc++.h>
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

const int N = 50011;
int a[N], n, m, cnt, head[N], sum;

struct node {
	int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
	e[++cnt].to = to;
	e[cnt].val = w;
	e[cnt].nxt = head[from];
	head[from] = cnt;
}

void dfs(int x, int fa) {
	for(int i = head[x]; i ; i = e[i].nxt) {
		int y = e[i].to;
		if(y == fa) continue;
		dfs(y, x);
		a[x] = e[i].val;
	}
}

int check(int k) {
	int t = 0, now = 1;
	for(int i = 1; i < n; i++) {
		if(now + a[i] >= k) {
			now = 0;
			t++;
		}
		else now += a[i];
	}
	return t >= m;
}

int main() {
	n = read(), m = read();
	for(int i = 1; i < n; i++) {
		int u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
		sum += w;
	}
	dfs(1, 0);
	int l = 1, r = sum, mid;
	while(l < r) {
		mid = (l + r) >> 1;
		if(check(mid)) l = mid;
		else r = mid - 1;
	}
	cout << l << '\n';
}

2、\(m = 1\) 求树的直径

#include <bits/stdc++.h>
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

const int N = 50011;
int a[N], n, m, cnt, head[N], sum, ans;

struct node {
	int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
	e[++cnt].to = to;
	e[cnt].val = w;
	e[cnt].nxt = head[from];
	head[from] = cnt;
}

int dfs(int x,int fa) {
	int sum1 = 0, sum2 = 0;
	for(int i = head[x]; i; i = e[i].nxt) {
		int y = e[i].to;
		if(y == fa) continue;
		sum2 = max(sum2, dfs(y, x) + e[i].val);
		if(sum2 > sum1) swap(sum1, sum2);
	}
	ans = max(ans, sum1 + sum2);
	return sum1;
}

int main() {
	n = read(), m = read();
	for(int i = 1; i < n; i++) {
		int u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
		sum += w;
	}
	dfs(1, 0);
	cout << ans << '\n';
}

3、\(a_i = 1\) 菊花图

#include <bits/stdc++.h>
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

const int N = 50011;
const int inf = 0x3f3f3f3f;
int a[N], n, m, cnt, head[N], sum, ans;

struct node {
	int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
	e[++cnt].to = to;
	e[cnt].val = w;
	e[cnt].nxt = head[from];
	head[from] = cnt;
}

bool cmp(int a, int b) {
	return a > b;
}

int main() {
	n = read(), m = read();
	for(int i = 1; i < n; i++) {
		int u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
		sum += w;
	}
	for(int i = head[1], y; i; i = e[i].nxt) {
		y = e[i].to;
		a[y - 1] = e[i].val;
	}
	sort(a + 1, a + n, cmp);
	int ans = inf;
	for(int i = 1; i <= m; i++)
		ans = min(ans, a[i] + a[2 * m - i + 1]);
	cout << ans << '\n';
	return 0;
}

4、混起来的部分分

#include <bits/stdc++.h>
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

const int N = 50011;
const int inf = 0x3f3f3f3f;
int a[N], n, m, cnt, head[N], sum, ans;

struct node {
	int to, nxt, val;
} e[N << 1];

inline void add(int from, int to, int w) {
	e[++cnt].to = to;
	e[cnt].val = w;
	e[cnt].nxt = head[from];
	head[from] = cnt;
}

namespace subtask1 {
	int a[N];
	void dfs(int x, int fa) {
		for(int i = head[x]; i ; i = e[i].nxt) {
			int y = e[i].to;
			if(y == fa) continue;
			dfs(y, x);
			a[x] = e[i].val;
		}
	}

	int check(int k) {
		int t = 0, now = 1;
		for(int i = 1; i < n; i++) {
			if(now + a[i] >= k) {
				now = 0;
				t++;
			} else now += a[i];
		}
		return t >= m;
	}

	void solve() {
		dfs(1, 0);
		int l = 1, r = sum, mid;
		while(l < r) {
			mid = (l + r + 1) >> 1;
			if(check(mid)) l = mid;
			else r = mid - 1;
		}
		cout << l << '\n';
	}
}

namespace subtask2 {
	int dfs(int x,int fa) {
		int sum1 = 0, sum2 = 0;
		for(int i = head[x]; i; i = e[i].nxt) {
			int y = e[i].to;
			if(y == fa) continue;
			sum2 = max(sum2, dfs(y, x) + e[i].val);
			if(sum2 > sum1) swap(sum1, sum2);
		}
		ans = max(ans, sum1 + sum2);
		return sum1;
	}
	void solve() {
		dfs(1, 0);
		cout << ans << '\n';
	}
}

namespace subtask3 {
	bool cmp(int a, int b) {
		return a > b;
	}
	void solve() {
		for(int i = head[1], y; i; i = e[i].nxt) {
			y = e[i].to;
			a[y - 1] = e[i].val;
		}
		sort(a + 1, a + n, cmp);
		int ans = inf;
		for(int i = 1; i <= m; i++)
			ans = min(ans, a[i] + a[2 * m - i + 1]);
		cout << ans << '\n';
	}
}

int main() {
	n = read(), m = read();
	int flag = 1, f = 1;
	for(int i = 1; i < n; i++) {
		int u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
		if(u != 1) flag = 0;
		if(v != u + 1) f = 0;
		sum += w;
	}
	if(flag) {
		subtask3::solve();
	}
	else if(f){
		subtask1::solve();
	}
	else {
		subtask2::solve();
	}
	return 0;
}

5、正解！！（\(multiset\)）

#include <bits/stdc++.h>
using namespace std;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

const int N = 50011;
const int inf = 0x3f3f3f3f;
int a[N], n, m, cnt, head[N], ans, up;

struct node {
	int to, nxt, val;
} e[N << 1];

multiset<int> s[N];
multiset<int>::iterator it;

inline void add(int from, int to, int w) {
	e[++cnt].to = to;
	e[cnt].val = w;
	e[cnt].nxt = head[from];
	head[from] = cnt;
}

int dfs(int x, int fa, int k) {
	s[x].clear();
	int w;
	for(int i =  head[x]; i; i = e[i].nxt) {
		int y = e[i].to;
		if(y == fa) continue;
		w = dfs(y, x, k) + e[i].val;
		if(w >= k) ans++;
		else s[x].insert(w);
	}
	int maxn = 0;
	while(!s[x].empty()) {
		if(s[x].size() == 1) {
			return max(maxn, *s[x].begin());
		}
		it = s[x].lower_bound(k - *s[x].begin());
		if(it == s[x].begin() && s[x].count(*it) == 1) it++;
		if(it == s[x].end()) {
			maxn = max(maxn, *s[x].begin());
			s[x].erase(s[x].find(*s[x].begin()));
		} else {
			ans++;
			s[x].erase(s[x].find(*it));
			s[x].erase(s[x].find(*s[x].begin()));
		}
	}
	return maxn;
}

int dfs1(int x,int fa) {
	int sum1 = 0, sum2 = 0;
	for(int i = head[x], y; i; i = e[i].nxt) {
		y=e[i].to;
		if(y == fa) continue;
		sum2 = max(sum2, dfs1(y, x) + e[i].val);
		if(sum1 < sum2) swap(sum1, sum2);
	}
	up = max(up, sum1 + sum2);
	return sum1;
}

int check(int k) {
	ans = 0;
	dfs(1, 0, k);
	if(ans >= m) return 1;
	return 0;
}

int main() {
	n = read(), m = read();
	for(int i = 1; i < n; i++) {
		int u = read(), v = read(), w = read();
		add(u, v, w);
		add(v, u, w);
	}
	dfs1(1, 0);
	int l = 1, r = up, mid;
	while(l < r) {
		mid = (l + r + 1) >> 1;
		if(check(mid)) l = mid;
		else r = mid - 1;
	}
	cout << l << '\n';
}