Problem G. Pandaria

給定一個(gè)有$n$條邊的無向連通圖，每條邊有對(duì)應(yīng)的邊權(quán)，每個(gè)點(diǎn)有一個(gè)顏色，

問從一個(gè)點(diǎn)出發(fā)，經(jīng)過不超過$w$的邊權(quán)，所能到達(dá)的點(diǎn)中，顏色出現(xiàn)次數(shù)做多且顏色編號(hào)最小的是什么顏色。

不超過某個(gè)權(quán)值所能到達(dá)的點(diǎn)，由此我們可以考慮建立升序$kruskal$重構(gòu)樹，然后從某個(gè)點(diǎn)倍增往上跳，直到不能跳為止，

這個(gè)時(shí)候我們所在的點(diǎn)的子樹所包含的點(diǎn)就是我們能夠到達(dá)的點(diǎn)了，

考慮用權(quán)值線段樹來維護(hù)每一個(gè)點(diǎn)所代表的子樹的信息，更新的時(shí)候我們只要往上進(jìn)行線段樹合并就行了。

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10;int n, nn, m, Q, ff[N], value[N], color[N], fa[N][21], ans[N];int root[N], ls[N << 5], rs[N << 5], maxn[N << 5], num;vector<int> G[N];struct Res {int u, v, w;void read() {scanf("%d %d %d", &u, &v, &w);}bool operator < (const Res &t) const {return w < t.w;} }edge[N];void push_up(int rt) {maxn[rt] = max(maxn[ls[rt]], maxn[rs[rt]]); }void update(int &rt, int l, int r, int x, int v) {if (!rt) {rt = ++num;}if (l == r) {maxn[rt] += v;return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], l, mid, x, v);}else {update(rs[rt], mid + 1, r, x, v);}push_up(rt); }int merge(int x, int y, int l, int r) {if (x == 0 || y == 0) {return x | y;}if (l == r) {maxn[x] += maxn[y];return x;}int mid = l + r >> 1;ls[x] = merge(ls[x], ls[y], l, mid);rs[x] = merge(rs[x], rs[y], mid + 1, r);push_up(x);return x; }int query(int rt, int l, int r) {if (l == r) {return l;}int mid = l + r >> 1;if (maxn[ls[rt]] == maxn[rt]) {return query(ls[rt], l, mid);}else {return query(rs[rt], mid + 1, r);} }void dfs(int rt, int f) {fa[rt][0] = f;for (int i = 1; i <= 20; i++) {fa[rt][i] = fa[fa[rt][i - 1]][i - 1];}if (rt <= n) {update(root[rt], 1, n, color[rt], 1);}for (int to : G[rt]) {if (to == f) {continue;}dfs(to, rt);root[rt] = merge(root[rt], root[to], 1, n);}ans[rt] = query(root[rt], 1, n); }int find(int rt) {return ff[rt] == rt ? rt : ff[rt] = find(ff[rt]); }void kruskal() {sort(edge + 1, edge + 1 + m);for (int i = 1; i < 2 * n; i++) {ff[i] = i;}for (int i = 1, cur = 1; i <= m && cur < n; i++) {int u = find(edge[i].u), v = find(edge[i].v);if (u ^ v) {nn++, cur++;ff[u] = ff[v] = nn;G[nn].push_back(u), G[nn].push_back(v);value[nn] = edge[i].w;if (u <= n) {value[u] = edge[i].w;}if (v <= n) {value[v] = edge[i].v;}}}for (int i = 1; i <= num; i++) {ls[i] = rs[i] = maxn[i] = 0;}for (int i = 1; i <= nn; i++) {root[i] = 0;}num = 0;dfs(nn, 0);for (int i = 1; i <= nn; i++) {G[i].clear();} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;scanf("%d", &T);for (int cas = 1; cas <= T; cas++) {printf("Case #%d:\n", cas);scanf("%d %d", &n, &m);nn = n;for (int i = 1; i <= n; i++) {scanf("%d", &color[i]);}for (int i = 1; i <= m; i++) {edge[i].read();}kruskal();scanf("%d", &Q);int res = 0;while (Q--) {int u, x;scanf("%d %d", &u, &x);u ^= res, x ^= res;for (int i = 20; i >= 0; i--) {if (fa[u][i] && value[fa[u][i]] <= x) {u = fa[u][i];}}res = ans[u];printf("%d\n", res);}}return 0; }

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