P3806 【模板】点分治1
解题思路：
点分治，我对于每次询问都直接计算...感觉复杂度挺大的..点分治处理出每个节点到根的距离，
排序后左右指针移动计算长度为k的路径数量
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
// clock_t c1 = clock();
// std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e5 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
struct Edge
{
int u, v, val, net;
} e[MAXN << 1];
int cnt = -1;
int head[MAXN];
int n, m, sum;
int k;
void add(int u, int v, int val)
{
e[++cnt].u = u;
e[cnt].v = v;
e[cnt].val = val;
e[cnt].net = head[u];
head[u] = cnt;
}
int mx[MAXN], vis[MAXN], sz[MAXN];
int rt;
int ans;
void dfs(int now, int fa)
{
sz[now] = 1, mx[now] = 0;
for (int i = head[now]; ~i; i = e[i].net)
{
int v = e[i].v;
if (vis[v] || v == fa)
continue;
dfs(v, now);
sz[now] += sz[v];
mx[now] = max(mx[now], sz[v]);
}
mx[now] = max(mx[now], sum - sz[now]);
if (mx[now] < mx[rt])
rt = now;
}
int tot;
int dis[MAXN];
void getdis(int now, int fa, int len)
{
for (int i = head[now]; ~i; i = e[i].net)
{
int v = e[i].v;
if (vis[v] || v == fa)
continue;
dis[++tot] = len + e[i].val;
getdis(v, now, dis[tot]);
}
}
int solve(int now, int len)
{
int ret = 0;
dis[tot = 1] = len;
getdis(now, -1, len);
int l = 1, r = tot;
sort(dis + 1, dis + 1 + tot);
while (l < r)
{
if (dis[l] + dis[r] == k)
{
l++, r--;
ret++;
}
else if (dis[l] + dis[r] < k)
l++;
else
r--;
}
// cout << ret << endl;
return ret;
}
void divide(int now)
{
vis[now] = 1;
ans += solve(now, 0);
for (int i = head[now]; ~i; i = e[i].net)
{
int v = e[i].v;
if (vis[v])
continue;
ans -= solve(v, e[i].val);
rt = 0;
sum = sz[v];
dfs(v, now);
divide(rt);
}
}
void init()
{
memset(head, -1, sizeof(head));
cnt = -1;
rt = 0;
mx[0] = inf;
}
int main()
{
scanf("%d%d", &n, &m);
init();
for (int i = 0; i < n - 1; i++)
{
int u, v, val;
scanf("%d%d%d", &u, &v, &val);
add(u, v, val);
add(v, u, val);
}
for (int i = 0; i < m; i++)
{
scanf("%d", &k);
memset(vis, 0, sizeof(vis));
ans = 0;
rt = 0;
sum = n;
dfs(1, -1);
divide(rt);
printf("%s\n", ans ? "AYE" : "NAY");
}
return 0;
}
P2634 [国家集训队]聪聪可可
解题思路：
在上题代码上改solve函数就差不多了,模3后到根结点路径长度为0,1,2的路径对于答案的贡献是\(dis[0]*dis[0]+2*dis[1]*dis[2]\)
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
// clock_t c1 = clock();
// std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 2e5 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
struct Edge
{
int u, v, val, net;
} e[MAXN << 1];
int cnt = -1;
int head[MAXN];
int n;
int sum;
void add(int u, int v, int val)
{
e[++cnt].u = u;
e[cnt].v = v;
e[cnt].val = val;
e[cnt].net = head[u];
head[u] = cnt;
}
int mx[MAXN], vis[MAXN], sz[MAXN];
int rt;
ll ans;
void dfs(int now, int fa)
{
sz[now] = 1, mx[now] = 0;
for (int i = head[now]; ~i; i = e[i].net)
{
int v = e[i].v;
if (vis[v] || v == fa)
continue;
dfs(v, now);
sz[now] += sz[v];
mx[now] = max(mx[now], sz[v]);
}
mx[now] = max(mx[now], sum - sz[now]);
if (mx[now] < mx[rt])
rt = now;
}
ll dis[4];
void getdis(int now, int fa, int len)
{
for (int i = head[now]; ~i; i = e[i].net)
{
int v = e[i].v;
if (vis[v] || v == fa)
continue;
dis[(len + e[i].val) % 3]++;
getdis(v, now, len + e[i].val);
}
}
ll solve(int now, int len)
{
dis[0] = dis[1] = dis[2] = 0;
dis[len % 3]++;
getdis(now, -1, len);
return dis[0] * dis[0] + 2 * dis[1] * dis[2];
}
void divide(int now)
{
vis[now] = 1;
ans += solve(now, 0);
for (int i = head[now]; ~i; i = e[i].net)
{
int v = e[i].v;
if (vis[v])
continue;
ans -= solve(v, e[i].val);
rt = 0;
sum = sz[v];
dfs(v, now);
divide(rt);
}
}
void init()
{
memset(vis, 0, sizeof(vis));
memset(head, -1, sizeof(head));
cnt = -1;
rt = 0;
mx[0] = inf;
}
ll GCD(ll a, ll b) { return b == 0 ? a : GCD(b, a % b); }
int main()
{
scanf("%d", &n);
init();
for (int i = 0; i < n - 1; i++)
{
int u, v, val;
scanf("%d%d%d", &u, &v, &val);
val %= 3;
add(u, v, val);
add(v, u, val);
}
sum = n;
dfs(1, -1);
divide(rt);
ll div = GCD(ans, 1LL * n * n);
printf("%lld/%lld\n", ans / div, 1LL * n * n / div);
// system("pause");
return 0;
}