【洛谷】P3128 [USACO15DEC] Max Flow P 的题解

【洛谷】P3128 [USACO15DEC] Max Flow P 的题解

题目传送门

题解

谔谔，LCA $+$ 树上差分，差点就被难倒了 qaq

今天就是 CSP 初赛了，祝大家也祝我自己 rp++！！！

其实是一道树上差分的板子题，先建树，对于一条路径 $w(u,v)$，将其权值 $+1$，然后再将它们的 LCA 和 LCA 的父亲各减去 $1$（LCA 用倍增比较方便），最后用 DFS 遍历整棵树统计和即可。

代码

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read() {
	register int x = 0, f = 1;
	register 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 x * f;
}
inline void write(int x) {
	if(x < 0) putchar('-'),x = -x;
	if(x > 9) write(x / 10);
	putchar(x % 10 + '0');
	return;
}
struct node {
	int to, next;
} edge[500005];
int fa[50005][30], head[500005], power[50005], depth[50005], lg[50005];
int n, k, ans = 0, tot = 0;
void add(int x, int y) {
	edge[++ tot].to = y;
	edge[tot].next = head[x];
	head[x] = tot;
}
void dfs(int now, int father) {
	fa[now][0] = father;
	depth[now] = depth[father] + 1;
	for(int i = 1; i <= lg[depth[now]]; ++i)
		fa[now][i] = fa[fa[now][i - 1]][i - 1];
	for(int i = head[now]; i; i = edge[i].next)
		if (edge[i].to != father) dfs(edge[i].to, now);
}
int LCA(int x, int y) {
	if(depth[x] < depth[y]) swap(x, y);
	while(depth[x] > depth[y]) x = fa[x][lg[depth[x] - depth[y]] - 1];
	if(x == y) return x;
	for(int k = lg[depth[x]] - 1; k >= 0; k --) {
		if(fa[x][k] != fa[y][k]) x = fa[x][k], y = fa[y][k];
	}
	return fa[x][0];
}
void find(int u, int father) {
	for(int i = head[u]; i; i = edge[i].next) {
		int to = edge[i].to;
		if(to == father) continue;
		find(to, u);
		power[u] += power[to];
	}
	ans = max(ans, power[u]);
}
int main() {
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	n = read(), k = read();
	int x, y;
	for(int i = 1; i <= n; i ++) {
		lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
	}
	for(int i = 1; i <= n - 1; i ++) {
		x = read(), y = read();
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	int s, t;
	for(int i = 1; i <= k; i ++) {
		s = read(), t = read(); 
		int ancestor = LCA(s, t);
		power[s] ++;
		power[t] ++;
		power[ancestor] --;
		power[fa[ancestor][0]] --;
	}
	find(1, 0);
	write(ans);
	return 0;
}