POJ 3585 Accumulation Degree

二次扫描与换根法

用于解决无根树,对于每一个节点作为根时都要统计

做法:

1.先以任意一个节点为根,做树形DP,保存每个节点的DP值

2.然后自上而下dfs,对于每个节点考虑以他为根的最大值

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 200505;
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9'){
		rv = (rv<<1) + (rv<<3) + c -'0';
		c = getchar();
	}
	return fh * rv;
}
int T, n, f[MAXN], c[MAXN], degree[MAXN], head[MAXN], nume, rot;
bool fff[MAXN];
struct edge{
	int to, nxt, flow;
}e[MAXN << 1];
void adde(int from, int to, int flow) {
	e[++nume].to = to;
	e[nume].flow = flow;
	e[nume].nxt = head[from];
	head[from] = nume;
}
void dfs1(int u) {
	fff[u] = 1;
	for(int i = head[u]; i ; i = e[i].nxt) {
		int v = e[i].to;
		if(fff[v]) continue;
		if(degree[v] == 1) {f[u] += e[i].flow;}
		else {dfs1(v); f[u] += min(f[v], e[i].flow);}
	}
}
void dfs2(int u) {
	fff[u] = 1;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(fff[v]) continue;
		c[v] = f[v];
		if(degree[u] == 1) c[v] += e[i].flow;
		else {
			c[v] += min(c[u] - min(f[u], e[i].flow), e[i].flow);
		}
		dfs2(v);
	}
}
int main() {
	T = init();
	while(T--) {
		n = init();
		memset(fff, 0, sizeof(fff));
		memset(f, 0, sizeof(f));
		memset(c, 0, sizeof(c));
		memset(degree, 0, sizeof(degree));
		memset(head, 0, sizeof(head));
		memset(e, 0, sizeof(e));
		nume = 0;
		for(int i = 1; i < n; i++) {
			int u = init(), v = init(), flow = init();
			adde(u, v, flow); adde(v, u, flow);
			degree[u]++; degree[v]++;
		}
		rot = 1;
		dfs1(rot);
		c[rot] = f[rot];
		memset(fff, 0, sizeof(fff));
		dfs2(rot);
		int ans = 0;
		for(int i = 1;i <= n; i++) ans = max(ans, c[i]);
		cout << ans << endl;
	}

	return 0;
}