版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/qq_39972971/article/details/85215029
【思路要点】
- 首先考虑某一种颜色,若该颜色在各连通块中的出现次数为 ,则该颜色对答案的贡献应为 ,因此我们只需要考虑各连通块各颜色出现次数平方的和即可。
- 考虑一棵树的做法,显然直接启发式合并 或是线段树合并即可计算每一棵子树中各颜色出现次数平方的和。
- 对原图建一棵圆方树,套用树的做法即可。
- 时间复杂度为 或 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 3e5 + 5; typedef long long ll; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct Inf { ll sum; map <int, int> cnt; void clear() { cnt.clear(); sum = 0; } void add(int x, int y) { int tmp = cnt[x]; sum -= 1ll * tmp * tmp; cnt[x] += y, tmp = cnt[x]; sum += 1ll * tmp * tmp; } } panorama, sub; struct Ans { ll sum, msum; map <int, int> cnt; int size() { return cnt.size(); } void clear() { cnt.clear(), sum = 0; msum = sub.sum; } void add(int x, int y) { int tmp = cnt[x]; sum -= 1ll * tmp * tmp; tmp = sub.cnt[x] - tmp; msum -= 1ll * tmp * tmp; cnt[x] += y, tmp = cnt[x]; sum += 1ll * tmp * tmp; tmp = sub.cnt[x] - tmp; msum += 1ll * tmp * tmp; } void merge(Ans &from) { for (map <int, int> :: iterator x = from.cnt.begin(); x != from.cnt.end(); x++) add((*x).first, (*x).second); } } res[MAXN]; int n, m, tot, timer, col[MAXN], dfn[MAXN], low[MAXN]; int top, Stack[MAXN]; ll psum, ans[MAXN]; vector <int> a[MAXN], b[MAXN]; void getans(int pos, int fa) { res[pos].clear(); if (pos <= n) res[pos].add(col[pos], 1); ans[pos] += sub.sum - 1; for (unsigned i = 0; i < b[pos].size(); i++) if (b[pos][i] != fa) { getans(b[pos][i], pos); ans[pos] -= res[b[pos][i]].sum; if (res[b[pos][i]].size() > res[pos].size()) swap(res[pos], res[b[pos][i]]); res[pos].merge(res[b[pos][i]]); } ans[pos] -= res[pos].msum; } void work(int pos) { sub.add(col[pos], 1); Stack[++top] = pos; dfn[pos] = low[pos] = ++timer; for (unsigned i = 0; i < a[pos].size(); i++) if (dfn[a[pos][i]] == 0) { work(a[pos][i]); chkmin(low[pos], low[a[pos][i]]); if (low[a[pos][i]] == dfn[pos]) { int tmp = pos; b[++tot].push_back(tmp); b[tmp].push_back(tot); while (tmp != a[pos][i]) { tmp = Stack[top--]; b[tot].push_back(tmp); b[tmp].push_back(tot); } } } else chkmin(low[pos], dfn[a[pos][i]]); } int main() { freopen("connection.in", "r", stdin); freopen("connection.out", "w", stdout); read(n), read(m), tot = n; for (int i = 1; i <= n; i++) { read(col[i]); panorama.add(col[i], 1); } for (int i = 1; i <= m; i++) { int x, y; read(x), read(y); a[x].push_back(y); a[y].push_back(x); } ll psum = panorama.sum; for (int i = 1; i <= n; i++) if (dfn[i] == 0) { sub.clear(); work(i); psum -= sub.sum; getans(i, 0); } for (int i = 1; i <= n; i++) printf("%lld\n", (ans[i] + psum) / 2); return 0; }