点分治学习笔记

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点分治学习笔记

简介

点分治主要解决树上路径问题，其主要思想是把一颗有根树以根为分治点分为一个森林（其实就是各个子树），解决经过当前根的路径后在子树里继续分治，从而将问题“分而治之”。

这里面，根的选择非常重要。为了保证复杂度，我们的分治点应该尽可能的“居中”，所以分治点一般选择正在处理的树的重心。

套路

  1. 找到当前树的重心作为根
  2. 解决通过这个根的路径的答案（一般有两种方法，一种是通过与子树容斥，一种是直接计算子树贡献）
  3. 递归解决子树

实现

求重心

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//root = 0, mxson[0] = INF, sum = n;
//root = 0, mxson[0] = INF, sum = size[edge[i].to];
void findRoot(int x, int f){
	size[x] = 1, mxson[x] = 0;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		findRoot(edge[i].to, x);
		size[x] += size[edge[i].to];
		mxson[x] = max(mxson[x], size[edge[i].to]);
	}
	mxson[x] = max(mxson[x], sum - size[x]);
	if(mxson[x] < mxson[root])	root = x;
}

size[]是子树大小，mxson[]是最大子树大小，root是重心，sum是当前整颗树的大小
注意每次 findRoot() 前要初始化 root 和 sum

分治计算

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void solve(int x){
	cal(x);//如果此处计算时将子树中一些不合法的路径的贡献也算进去了，那么需要容斥，即在下方*处减掉子树贡献；如果不容斥，就直接利用每颗子树的信息计算答案
	vis[x] = 1;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		cal(edge[i].to);//*
		root = 0, sum = size[edge[i].to], mxson[0] = INF;
		findRoot(edge[i].to, 0);
		solve(root);
	}
}

vis[]标记此点是否计算过，cal()计算以x为根、经过根的路径的答案

注意事项

不要再分治时用memset O(n) 地进行初始化，否则点分治好不容易保证的复杂度就被毁了。

练习

poj1741 tree

题意
给一棵树，边有边权，问两点之间的距离小于等于K的点对有多少个。
题解
点分治时用容斥做：计算以x为根的子树时直接将求得的dis排序后O(n)求答案，然后再减去每个子树中被统计了的不合法答案
Code

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# include<cstdio>
# include<cstring>
# include<algorithm>

using namespace std;

const int INF = 1e9;
const int N = 10005;
int n, k, u, v, p;

struct Edge{
	int nxt, to, dis;
}edge[N<<1];
int head[N], edgeNum;
void addEdge(int from, int to, int dis){
	edge[++edgeNum].nxt = head[from];
	edge[edgeNum].to = to;
	edge[edgeNum].dis = dis;
	head[from] = edgeNum;
}

bool vis[N];
int root, sum, size[N], mxson[N];
void findRoot(int x, int f){
	size[x] = 1, mxson[x] = 0;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(edge[i].to == f || vis[edge[i].to])	continue;
		findRoot(edge[i].to, x);
		size[x] += size[edge[i].to];
		mxson[x] = max(mxson[x], size[edge[i].to]);
	}
	mxson[x] = max(mxson[x], sum - size[x]);
	if(mxson[x] < mxson[root])	root = x;
}

int dis[N], ans;
void getDis(int x, int f, int d){
	dis[++dis[0]] = d;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(edge[i].to == f || vis[edge[i].to])	continue;
		getDis(edge[i].to, x, d + edge[i].dis);
	}
}

int cal(int x, int d){
	int res = 0;
	for(int i = 1; i <= dis[0]; i++)	dis[i] = 0;
	dis[0] = 0;
	getDis(x, 0, d);
	sort(dis+1, dis+dis[0]+1);
	int l = 1, r = dis[0];
	while(l < r){
		if(dis[l] + dis[r] <= k)	res += r - l, l++;
		else	r--;
	}
	return res;
}

void solve(int x){
	ans += cal(x, 0);
	vis[x] = 1;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		ans -= cal(edge[i].to, edge[i].dis);
		root = 0, mxson[0] = INF, sum = size[edge[i].to];
		findRoot(edge[i].to, 0);
		solve(root);
	}
}

void init(){
	memset(edge, 0, sizeof edge);
	memset(head, 0, sizeof head);
	edgeNum = 0;
	root = sum = ans = 0;
	memset(size, 0, sizeof size);
	memset(mxson, 0, sizeof mxson);
	memset(vis, 0, sizeof vis);
}

int main(){
	while(1){
		scanf("%d%d", &n, &k);
		if(n == 0 && k == 0)	break;
		init();
		for(int i = 1; i < n; i++){
			scanf("%d%d%d", &u, &v, &p);
			addEdge(u, v, p);
			addEdge(v, u, p);
		}
		sum = n, root = 0, mxson[0] = INF;
		findRoot(1, 0);
		solve(root);
		printf("%d\n", ans);
	}
	return 0;
}

luogu3806 【模板】点分治1

题意
给定一棵有n个点的树，多次询问树上距离为k的点对是否存在。
题解
和上一题差不多，也是容斥，只不过我们把所有k的答案一次性求出来，每次询问O(1)回答。
Code

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# include<map>
# include<cstdio>
# include<cstring>
# include<algorithm>

using namespace std;

const int INF = 1e9;
const int N = 10005;
int n, m, k[105], u, v, p;

struct Edge{
	int nxt, to, dis;
}edge[N<<1];
int head[N], edgeNum;
void addEdge(int from, int to, int dis){
	edge[++edgeNum].nxt = head[from];
	edge[edgeNum].to = to;
	edge[edgeNum].dis = dis;
	head[from] = edgeNum;
}

int size[N], mxson[N], dis[N], root, sum;
bool vis[N];
map<int, int> cnt;
void findRoot(int x, int f){
	size[x] = 1, mxson[x] = 0;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		findRoot(edge[i].to, x);
		size[x] += size[edge[i].to];
		mxson[x] = max(mxson[x], size[edge[i].to]);
	}
	mxson[x] = max(mxson[x], sum - size[x]);
	if(mxson[x] < mxson[root])	root = x;
}

void getDis(int x, int f, int d){
	dis[++dis[0]] = d;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(edge[i].to == f || vis[edge[i].to])	continue;
		getDis(edge[i].to, x, d + edge[i].dis);
	}
}

void cal(int x, int d, int fl){
	for(int i = 1; i <= dis[0]; i++)	dis[i] = 0;
	dis[0] = 0;
	getDis(x, 0, d);
	sort(dis+1, dis+dis[0]+1);
	for(int i = 1; i <= dis[0]; i++){
		for(int j = 1; j <= m; j++){
			if(dis[i] + dis[i] > k[j])	continue;
			int l = lower_bound(dis+i, dis+dis[0]+1, k[j]-dis[i]) - dis;
			if(dis[l] + dis[i] != k[j])	continue;
			int r = upper_bound(dis+i, dis+dis[0]+1, k[j]-dis[i]) - dis;
			cnt[k[j]] += (r - l) * fl;
		}
	}
}

void solve(int x){
	cal(x, 0, 1);
	vis[x] = 1;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		cal(edge[i].to, edge[i].dis, -1);
		root = 0, sum = size[edge[i].to], mxson[0] = INF;
		findRoot(edge[i].to, 0);
		solve(root);
	}
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i < n; i++){
		scanf("%d%d%d", &u, &v, &p);
		addEdge(u, v, p);
		addEdge(v, u, p);
	}
	for(int i = 1; i <= m; i++)
		scanf("%d", &k[i]);
	root = 0, sum = n, mxson[0] = INF;
	findRoot(1, 0);
	solve(root);
	for(int i = 1; i <= m; i++)
		puts(cnt[k[i]] > 0 ? "AYE" : "NAY");
	return 0;
}

[国家集训队] 聪聪可可

题意
求边权和是3的倍数的点对个数
题解
思路和上面两道题大同小异，而且更简单了：不用对dis排序，只需记录下距当前根dis为0,1,2的点的个数(cnt)，则答案就是$cnt[0] * cnt[0] + cnt[1] * cnt[2] * 2$
当然这样做也要容斥
Code

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# include<cstdio>
# include<cstring>
# include<algorithm>

using namespace std;

const int N = 20005;
const int INF = 1e9;
int n, u, v, p;

struct Edge{
	int nxt, to, dis;
}edge[N<<1];
int head[N], edgeNum;
void addEdge(int from, int to, int dis){
	edge[++edgeNum].nxt = head[from];
	edge[edgeNum].to = to;
	edge[edgeNum].dis = dis;
	head[from] = edgeNum;
}

int root, mxson[N], sum, size[N];
bool vis[N];
void findRoot(int x, int f){
	size[x] = 1, mxson[x] = 0;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		findRoot(edge[i].to, x);
		size[x] += size[edge[i].to];
		mxson[x] = max(mxson[x], size[edge[i].to]);
	}
	mxson[x] = max(mxson[x], sum - size[x]);
	if(mxson[x] < mxson[root])	root = x;
}

int cnt[3];
void getDis(int x, int f, int d){
	cnt[d%3]++;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		getDis(edge[i].to, x, (d + edge[i].dis) % 3);
	}
}

int cal(int x, int d){
	cnt[0] = cnt[1] = cnt[2] = 0;
	getDis(x, 0, d);
	return cnt[0] * cnt[0] + cnt[1] * cnt[2] * 2;
}

int ans;
void solve(int x){
	ans += cal(x, 0);
	vis[x] = 1;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		ans -= cal(edge[i].to, edge[i].dis % 3);
		root = 0, sum = size[edge[i].to], mxson[0] = INF;
		findRoot(edge[i].to, 0);
		solve(root);
	}
}

int gcd(int a, int b){
	return b == 0 ? a : gcd(b, a % b);
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i < n; i++){
		scanf("%d%d%d", &u, &v, &p);
		addEdge(u, v, p % 3);
		addEdge(v, u, p % 3);
	}
	root = 0, mxson[0] = INF, sum = n;
	findRoot(1, 0);
	solve(root);
	int g = gcd(ans, n*n);
	printf("%d/%d", ans / g,  n * n / g);
	return 0;
}

[IOI]Race

题意
给一棵树，每条边有权。求一条简单路径，权值和等于 K ，且边的数量最小。输出最小边数
题解
发现这道不能容斥…所以我们想办法通过子树信息直接计算经过分治点的路径的答案
记$tmp[i]$为当前子树中，路径长为$i$的最小边数，于是对于当前根$x$，我们每次遍历它的子树，先用$tmp[]$和正在遍历的子树更新答案（代码中的 $updAns()$ 函数），再用正在遍历的这颗子树更新$tmp[]$（代码中的 $updTmp()$ 函数），这样就保证了不会把不合法的路径算进来
Code

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# include<cstdio>
# include<cstring>
# include<algorithm>

using namespace std;

const int INF = 1e9;
const int N = 200005;
int n, k, u, v, p;

struct Edge{
	int nxt, to, dis;
}edge[N<<1];
int head[N], edgeNum;
void addEdge(int from, int to, int dis){
	edge[++edgeNum].nxt = head[from];
	edge[edgeNum].to = to;
	edge[edgeNum].dis = dis;
	head[from] = edgeNum;
}

int root, mxson[N], size[N], sum;
bool vis[N];
void findRoot(int x, int f){
	size[x] = 1, mxson[x] = 0;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		findRoot(edge[i].to, x);
		size[x] += size[edge[i].to];
		mxson[x] = max(mxson[x], size[edge[i].to]);
	}
	mxson[x] = max(mxson[x], sum - size[x]);
	if(mxson[x] < mxson[root])	root = x;
}

int ans = INF, tmp[1000005];
void updTmp(int x, int f, int dis, int d){
	if(dis <= k)	tmp[dis] = min(tmp[dis], d);
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		updTmp(edge[i].to, x, dis + edge[i].dis, d + 1);
	}
}
void updAns(int x, int f, int dis, int d){
	if(dis <= k)	ans = min(ans, d + tmp[k-dis]);
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		updAns(edge[i].to, x, dis + edge[i].dis, d + 1);
	}
}
void clearTmp(int x, int f, int dis){
	if(dis <= k)	tmp[dis] = INF;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to] || edge[i].to == f)	continue;
		clearTmp(edge[i].to, x, dis + edge[i].dis);
	}
}

void solve(int x){
	vis[x] = 1, tmp[0] = 0;
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		updAns(edge[i].to, x, edge[i].dis, 1);
		updTmp(edge[i].to, x, edge[i].dis, 1);
	}
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		clearTmp(edge[i].to, x, edge[i].dis);
	}
	for(int i = head[x]; i; i = edge[i].nxt){
		if(vis[edge[i].to])	continue;
		root = 0, sum = size[edge[i].to];
		findRoot(edge[i].to, 0);
		solve(root);
	}
}

int main(){
	scanf("%d%d", &n, &k);
	for(int i = 1; i < n; i++){
		scanf("%d%d%d", &u, &v, &p);
		addEdge(u+1, v+1, p);
		addEdge(v+1, u+1, p);
	}
	for(int i = 0; i <= k; i++)	tmp[i] = INF;
	root = 0, mxson[0] = INF, sum = n;
	findRoot(1, 0);
	solve(root);
	if(ans == INF)	puts("-1");
	else	printf("%d\n", ans);
	return 0;
}

—— 完 ——