短暂回归的省选练习题单

震惊，蒟蒻居然也能去省选，短暂回归 OI!
2020/5/15: 震惊，蒟蒻居然能混到决赛三等奖！

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题单

网络流： https://www.luogu.com.cn/training/65946#information
树链剖分： https://www.luogu.com.cn/training/65955#information
树的直径：https://www.luogu.com.cn/training/67339#information

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网络流答案：

P3171 吞吐量

跑最短路建图，然后拆点即可

#include <bits/stdc++.h>
using namespace std;
#define N 1000010
#define ll long long
const ll inf = 1e16; 

template <class T>
inline void read(T& a){
	T x = 0, s = 1;
	char c = getchar();
	while(!isdigit(c)){ if(c == '-') s = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + (c ^ '0'); c = getchar(); }
	a = x * s;
	return ;
}

struct node{
	int v, next; 
	ll w; 
} t[N << 1];
int f[N]; 
int cur[N]; 

int bian = -1;
inline void add(int u, int v, ll w){
	t[++bian] = (node){v, f[u], w}, f[u] = bian;
	t[++bian] = (node){u, f[v], 0}, f[v] = bian; 
	return ; 
}

int n, m; 

ll G[1010][1010];  

ll dis[N]; 
ll c[N]; 
bool vis[N]; 

void spfa(){
	queue <int> q; 
	memset(dis, 0x3f, sizeof(dis)); 
	memset(vis, 0, sizeof(vis)); 
	
	q.push(1); vis[1] = 1; dis[1] = 1;  
	while(!q.empty()){
		int now = q.front(); q.pop();
		vis[now] = 0; 
		for(int v = 1; v <= n; v++){
			if(~G[now][v]){
				if(dis[v] > dis[now] + G[now][v]){
					dis[v] = dis[now] + G[now][v]; 
					if(!vis[v]) q.push(v), vis[v] = 1; 
				}
			}
		}
	}
	return ; 
}

int deth[N]; 

bool bfs(int S, int T){
	queue <int> q; 
	memset(deth, 0, sizeof(deth)); 
	q.push(S), deth[S] = 1; 
	while(!q.empty()){
		int now = q.front(); q.pop();
		for(int i = f[now]; ~i; i = t[i].next){
			int v = t[i].v;
			if(!deth[v] && t[i].w != 0)
				deth[v] = deth[now] + 1, q.push(v); 
		}
	}
	return deth[T];  // 能到达汇点 
}

ll dfs(int now, int T, ll flow){
	if(now == T || !flow) return flow; 
	for(int& i = cur[now]; ~i; i = t[i].next){
		int v = t[i].v;
		if(deth[v] == deth[now] + 1 && t[i].w != 0){
			ll di = dfs(v, T, min(t[i].w, flow)); 
			if(di > 0){
				t[i].w -= di;
				t[i^1].w += di;
				return di; 
			}
		}
	}
	return 0;
}

ll Dicnic(int S, int T){
	ll sum = 0;
	while(bfs(S, T)){
		memcpy(cur, f, sizeof(cur));
		ll tmp = 0; 
		while((tmp = dfs(S, T, inf))) sum += tmp; 
	}
	return sum; 
}

int main(){
//	freopen("hh.txt", "r", stdin);  
	read(n), read(m);
	memset(G, -1, sizeof(G)); 
	for(int i = 1; i <= m; i++){
		int x, y; ll w; 
		read(x), read(y), read(w); 
		if(G[x][y] == -1) G[x][y] = G[y][x] = w; 
		else G[x][y] = G[y][x] = min(G[x][y], w); 
	}
	for(int i = 1; i <= n; i++) read(c[i]); 
	spfa(); 
	memset(f, -1, sizeof(f)); 
	int S = 0, T = n + n + 1; 
	add(S, 1, inf); add(n << 1, T, inf); 
	for(int i = 1; i <= n; i++){
		if(i != 1 && i != n) add(i, i + n, c[i]);
		else add(i, i + n, inf); 
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			if(i != j)
				if(dis[i] + G[i][j] == dis[j] && ~G[i][j]) add(i + n, j, inf); 
		}
	}
	ll ans = Dicnic(S, T);
	cout << ans << endl; 
	return 0;
}

树链剖分答案：

P3258 松鼠的新家

其实直接树上差分即可，但还是练习一下树剖。

#include <bits/stdc++.h>
using namespace std;
#define N 1000010
#define ll long long

template <class T>
inline void read(T& a){
	T x = 0, s = 1;
	char c = getchar();
	while(!isdigit(c)){ if(c == '-') s = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + (c ^ '0'); c = getchar(); }
	a = x * s;
	return ;
}

struct node{
	int v, next;
} t[N << 1];
int f[N];

int bian = 0;
inline void addedge(int u, int v){
	t[++bian] = (node){v, f[u]}, f[u] = bian;
	return ; 
}

int n; 
int a[N]; 

int top[N], fa[N], deth[N], son[N], siz[N]; 
int dfn[N], id = 0; 

#define v t[i].v

void dfs1(int now, int father){
	fa[now] = father;
	deth[now] = deth[father] + 1; 
	siz[now] = 1; 
	for(int i = f[now]; i; i = t[i].next){
		if(v != father){
			dfs1(v, now); 
			siz[now] += siz[v];
			if(siz[v] > siz[son[now]])
				son[now] = v; 
		}
	}
	return ; 
}

void dfs2(int now, int tp){
	top[now] = tp;
	dfn[now] = ++id; 
	if(!son[now]) return ;
	dfs2(son[now], tp);
	for(int i = f[now]; i; i = t[i].next){
		if(v != fa[now] && v != son[now])
			dfs2(v, v); 
	}
	return ; 
}

#undef v

struct Segment_tree{        /*完整版线段树练习*/
	
	private:
	
	struct node{
		int w, add;
	} e[N << 2];
	
	#define lson (o<<1)
	#define rson (o<<1|1)
	
	inline void pushdown(int o, int l, int r){
		int mid = l + r >> 1; 
		e[lson].add += e[o].add;
		e[rson].add += e[o].add;
		e[lson].w += (e[o].add * (mid - l + 1)); 
		e[rson].w += (e[o].add * (r - mid)); 
		e[o].add = 0; 
		return ; 
	} 
	
	inline void pushup(int o){
		e[o].w = e[lson].w + e[rson].w;
		return ; 
	}
	
	public:
/*	
	inline void build(int o, int l, int r){
		e[o].w = e[o].add = 0;
		if(l == r) return ;
		int mid = l + r >> 1; 
		build(lson, l, mid); 
		build(rson, mid + 1, r);
		pushup(o); 
	}
	此题中不需要，仅供练习 
*/
	
	void update(int o, int l, int r, int in, int end, int k){
		if(l > end || r < in) return ; 
		if(l >= in && r <= end){
			e[o].add += k;
			e[o].w += (r - l + 1) * k;
			return ; 
		}
		pushdown(o, l, r);
		int mid = l + r >> 1; 
		update(lson, l, mid, in, end, k);
		update(rson, mid + 1, r, in, end, k);
		pushup(o);
		return ; 
	}
	
	int query(int o, int l, int r, int x){
		if(l == r && l == x) return e[o].w; 
		pushdown(o, l, r);
		int mid = l + r >> 1; 
		if(x <= mid) return query(lson, l, mid, x);
		else if(x > mid) return query(rson, mid + 1, r, x); 
		return 0; 
	}
	
} tree;

void jia(int x, int y){
	while(top[x] != top[y]){
		if(deth[top[x]] < deth[top[y]]) swap(x, y); 
		tree.update(1, 1, n, dfn[top[x]], dfn[x], 1); 
		x = fa[top[x]]; 
	}
	if(deth[x] > deth[y]) swap(x, y);  
	tree.update(1, 1, n, dfn[x], dfn[y], 1); 
	return ; 
}

int main(){
//	freopen("hh.txt", "r", stdin); 
	read(n);
	for(int i = 1; i <= n; i++) read(a[i]);
	for(int i = 1; i < n; i++){
		int x, y; 
		read(x), read(y);
		addedge(x, y); addedge(y, x); 
	}
	dfs1(1, 0);
	dfs2(1, 0); 
	
//	tree.build(1, 1, n);   此题中不需要，仅供练习 
	
	for(int i = 1; i < n; i++){
		jia(a[i], a[i+1]); 
		tree.update(1, 1, n, dfn[a[i+1]], dfn[a[i+1]], -1);  // 注意，a[i+1] 会重复加一，应减掉 
	}
	for(int i = 1; i <= n; i++)
		printf("%d
", tree.query(1, 1, n, dfn[i])); 
	return 0;
}

树的直径答案

CF455C / P2195 HXY造公园

用并查集维护森林编号。然后可以发现，操作一可以 (O(N)) 预处理。操作二，有式子 (ans(new) = max(ans[x], ans[y], [(ans[x] + 1) / 2 + (ans[y] + 1) / 2 + 1]))

/*  2021.4.10 0:25 
for both problems:
	luogu: P2195 HXY造公园
	CF455C: Civilization
*/

#include <bits/stdc++.h>
using namespace std;
#define N 1000010
#define ll long long
const int inf = -2e9; 

template <class T>
inline void read(T& a){
	T x = 0, s = 1;
	char c = getchar();
	while(!isdigit(c)){ if(c == '-') s = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + (c ^ '0'); c = getchar(); }
	a = x * s;
	return ;
}

struct node{
	int v, next;
} t[N << 1];
int f[N];

int bian = 0;
inline void add(int u, int v){
	t[++bian] = (node){v, f[u]}, f[u] = bian;
	return ; 
}

/* 并查集  */

int ffa[N];

int find(int x){
	return x == ffa[x] ? x : ffa[x] = find(ffa[x]); 
}

inline void merge(int u, int v){
	int fau = find(u), fav = find(v); 
	ffa[fau] = fav;
	return ; 
}

/* 并查集 end */

int vis[N]; 
int dis[N]; 
int ans[N]; 
int find_root; 
int find_maxn = 0; 

int n, m, T; 

void dfs(int now, int rt, int dis){
	if(vis[now] == rt) return ;
	vis[now] = rt; 
	if(dis > find_maxn){
		find_root = now; 
		find_maxn = dis; 
	}
	if(now != rt) merge(now, rt); 
	for(int i = f[now]; i; i = t[i].next){
		int v = t[i].v;
		dfs(v, rt, dis + 1); 
	}
	return ; 
}

void get_root(){     // 寻找每棵树的直径 
	for(int i = 1; i <= n; i++)
		ffa[i] = i;     // 并查集初始化 
	memset(vis, 0, sizeof(vis)); 
	for(int i = 1; i <= n; i++){
		if(!vis[i]){
			find_root = i;  
			find_maxn = 0; 
			dfs(i, i, 0); 
			find_maxn = 0; 
			dfs(find_root, find_root, 0); 
			ans[find(i)] = find_maxn; 
		}
	}
	return ; 	
}

int main(){
	read(n), read(m), read(T);
	for(int i = 1; i <= m; i++){
		int x, y;
		read(x), read(y); 
		add(x, y); add(y, x); 
	}
	get_root();   // 找到每棵树的直径并记录 
	
	while(T--){
		int opt = 0, x = 0, y = 0;
		read(opt);
		switch (opt){
			case 1:
				read(x);
				printf("%d
", ans[find(ffa[x])]); 
				break; 
			case 2:
				read(x), read(y);
				x = find(x), y = find(y);
				if(x == y) break ;  
				merge(x, y);
				ans[find(x)] = max(ans[x], max(ans[y], (ans[x] + 1) / 2 + (ans[y] + 1) / 2 + 1)); 
				break ; 
		}
	}
	return 0;
}