题面
\(\text{Solution:}\)
有些题目不仅让我们做树型 \(\text{dp}\) ,而且还让我们换每个根分别做一次, 然后这样就愉快的 \(\text{TLE}\) 了,所以我们要用一种方法快速知道所有根的答案。
二次扫描与换根法:
就是先选任意点作根做一遍 \(\text{dp}\) ,求出相关信息,然后再从根往下 \(\text{dfs}\) ,对每一个节点往下走之前进行自顶向下的推导,计算出 "换根" 后的解。
就这题而言就是用父亲的换根后的答案来跟新自己换根前的答案,一般是 换根后父亲的答案+自己换根前的答案-自己对父亲换根后的贡献。
#include <set>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <assert.h>
#include <algorithm>
using namespace std;
#define fir first
#define sec second
#define pb push_back
#define mp make_pair
#define LL long long
#define INF (0x3f3f3f3f)
#define mem(a, b) memset(a, b, sizeof (a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(x) cout << #x << " = " << x << endl
#define travle(i, x) for (register int i = head[x]; i; i = nxt[i])
#define For(i, a, b) for (register int (i) = (a); (i) <= (b); ++ (i))
#define Forr(i, a, b) for (register int (i) = (a); (i) >= (b); -- (i))
#define file(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)
#define ____ debug("go\n")
namespace io {
static char buf[1<<21], *pos = buf, *end = buf;
inline char getc()
{ return pos == end && (end = (pos = buf) + fread(buf, 1, 1<<21, stdin), pos == end) ? EOF : *pos ++; }
inline int rint() {
register int x = 0, f = 1;register char c;
while (!isdigit(c = getc())) if (c == '-') f = -1;
while (x = (x << 1) + (x << 3) + (c ^ 48), isdigit(c = getc()));
return x * f;
}
inline LL rLL() {
register LL x = 0, f = 1; register char c;
while (!isdigit(c = getc())) if (c == '-') f = -1;
while (x = (x << 1ll) + (x << 3ll) + (c ^ 48), isdigit(c = getc()));
return x * f;
}
inline void rstr(char *str) {
while (isspace(*str = getc()));
while (!isspace(*++str = getc()))
if (*str == EOF) break;
*str = '\0';
}
template<typename T>
inline bool chkmin(T &x, T y) { return x > y ? (x = y, 1) : 0; }
template<typename T>
inline bool chkmax(T &x, T y) { return x < y ? (x = y, 1) : 0; }
}
using namespace io;
const int N = 2e5 + 2;
int n, T;
int tot, head[N], ver[N<<1], nxt[N<<1], edge[N<<1];
int D[N], F[N], in[N];
void add(int u, int v, int w)
{ ver[++tot] = v, edge[tot] = w, nxt[tot] = head[u], head[u] = tot; }
void DFS(int u, int f)
{
D[u] = 0;
for (int i = head[u]; i; i = nxt[i])
if (ver[i] != f)
{
DFS(ver[i], u);
int v = ver[i];
if (in[v] == 1) D[u] += edge[i];
else D[u] += min(D[v], edge[i]);
}
}
void DP(int u, int f)
{
for (int i = head[u]; i; i = nxt[i])
if (ver[i] != f)
{
int v = ver[i];
if (in[u] == 1) F[v] = D[v] + edge[i];
else F[v] = D[v] + min(F[u] - min(edge[i], D[v]), edge[i]);
DP(v, u);
}
}
int main() {
T = rint();
while (T --)
{
tot = 0; mem(head, 0); mem(in, 0); mem(D, 0); mem(F, 0);
n = rint();
for (int i = 1; i < n; ++ i)
{
int u = rint(), v = rint(), w = rint();
add(u, v, w);
add(v, u, w);
in[u] ++, in[v] ++;
}
DFS(1, 0);
F[1] = D[1];
DP(1, 0);
for (int i = 1; i <= n; ++ i)
chkmax(F[1], F[i]);
printf("%d\n", F[1]);
}
}
原文地址:https://www.cnblogs.com/HNYLMSTea/p/10505895.html
时间: 2024-11-05 22:47:00