【LG4185】[USACO18JAN]MooTube
题面
题解
先将所有操作和询问离线
然后按照边权从大到小将操作和询问排序
利用(two;pointers),每次扫到一个询问,将边权大于等于它的边的两点全部都并起来
因为边权大的满足,那么边权小的一定也能满足
对于每个询问,直接查它联通块的(size)即可
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
return w * data;
}
const int MAX_N = 1e5 + 5;
struct Query { int k, v, id; } q[MAX_N];
struct Edge { int u, v, w; } e[MAX_N];
bool operator < (const Query &l, const Query &r) { return l.k > r.k; }
bool operator < (const Edge &l, const Edge &r) { return l.w > r.w; }
int N, Q, fa[MAX_N], size[MAX_N], ans[MAX_N];
int getf(int x) { return (x == fa[x]) ? x : (fa[x] = getf(fa[x])); }
void unite(int x, int y) { x = getf(x), y = getf(y); if (x != y) fa[x] = y, size[y] += size[x]; }
int main () {
N = gi(), Q = gi();
for (int i = 1; i < N; i++) e[i].v = gi(), e[i].u = gi(), e[i].w = gi();
for (int i = 1; i <= Q; i++) q[i].k = gi(), q[i].v = gi(), q[i].id = i;
sort(&e[1], &e[N]); sort(&q[1], &q[Q + 1]);
for (int i = 1; i <= N; i++) fa[i] = i, size[i] = 1;
for (int i = 1, j = 1; i <= Q; i++) {
while (e[j].w >= q[i].k && j < N) unite(e[j].u, e[j].v), ++j;
ans[q[i].id] = size[getf(q[i].v)] - 1;
}
for (int i = 1; i <= Q; i++) printf("%d
", ans[i]);
return 0;
}