LCT維護生成樹

先按照a的權值把邊排序，離線維護b的最小生成樹。

將a排序后，依次動態加邊，我們只需要關注b的值。要保證1-n花費最少，兩點間的b值肯定是越小越好，所以我們可以考慮以b為關鍵字維護最小生成樹。

對于新加的邊b，如果1-n已經聯通，需要更新答案

#include <bits/stdc++.h> #define INF 0x3f3f3f3f #define full(a, b) memset(a, b, sizeof a) using namespace std; typedef long long ll; inline int lowbit(int x){ return x & (-x); } inline int read(){int X = 0, w = 0; char ch = 0;while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();return w ? -X : X; } inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; } inline int lcm(int a, int b){ return a / gcd(a, b) * b; } template<typename T> inline T max(T x, T y, T z){ return max(max(x, y), z); } template<typename T> inline T min(T x, T y, T z){ return min(min(x, y), z); } template<typename A, typename B, typename C> inline A fpow(A x, B p, C lyd){A ans = 1;for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;return ans; } const int N = 400005; int n, m, tot, ans, mx[N], fa[N], w[N], ch[N][2], rev[N], id[N], st[N]; struct Edge {int v, u, a, b;bool operator < (const Edge &rhs) const {return a < rhs.a;} } e[N];int newNode(int v){++tot;w[tot] = mx[tot] = v, id[tot] = tot;fa[tot] = ch[tot][0] = ch[tot][1] = 0;return tot; }bool isRoot(int x){return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }void reverse(int x){rev[x] ^= 1, swap(ch[x][0], ch[x][1]); }void push_up(int x){int l = ch[x][0], r = ch[x][1];mx[x] = w[x], id[x] = x;if(mx[l] > mx[x]) mx[x] = mx[l], id[x] = id[l];if(mx[r] > mx[x]) mx[x] = mx[r], id[x] = id[r]; }void push_down(int x){if(rev[x]){rev[x] ^= 1;if(ch[x][0]) reverse(ch[x][0]);if(ch[x][1]) reverse(ch[x][1]);} }void rotate(int x){int y = fa[x], z = fa[y], p = (ch[y][1] == x) ^ 1;ch[y][p^1] = ch[x][p], fa[ch[x][p]] = y;if(!isRoot(y)) ch[z][ch[z][1] == y] = x;fa[x] = z, fa[y] = x, ch[x][p] = y;push_up(y), push_up(x); }void splay(int x){int pos = 0; st[++pos] = x;for(int i = x; !isRoot(i); i = fa[i]) st[++pos] = fa[i];while(pos) push_down(st[pos--]);while(!isRoot(x)){int y = fa[x], z = fa[y];if(!isRoot(y)) (ch[y][0] == x) ^ (ch[z][0] == y) ? rotate(x) : rotate(y);rotate(x);}push_up(x); }void access(int x){for(int p = 0; x; p = x, x = fa[x])splay(x), ch[x][1] = p, push_up(x); }void makeRoot(int x){access(x), splay(x), reverse(x); }void link(int x, int y){makeRoot(x);fa[x] = y; }int findRoot(int x){access(x), splay(x);while(ch[x][0]) push_down(x), x = ch[x][0];splay(x);return x; }void split(int x, int y){makeRoot(x), access(y), splay(y); }bool isConnect(int x, int y){makeRoot(x);return findRoot(y) == x; }int main(){ans = INF;n = read(), m = read();for(int i = 1; i <= n; i ++) newNode(0);for(int i = 0; i < m; i ++){e[i].u = read(), e[i].v = read(), e[i].a = read(), e[i].b = read();}sort(e, e + m);for(int i = 0; i < m; i ++){int u = e[i].u, v = e[i].v, t = newNode(e[i].b);if(!isConnect(u, v)) link(u, t), link(t, v);else{split(u, v);if(e[i].b > mx[v]) continue;int tmp = id[v]; splay(tmp);fa[ch[tmp][0]] = fa[ch[tmp][1]] = 0;ch[tmp][0] = ch[tmp][1] = 0;link(u, t), link(t, v);}if(isConnect(1, n)){split(1, n);ans = min(ans, mx[n] + e[i].a);}}printf(ans == INF ? "-1\n" : "%d\n", ans);return 0; }

轉載于:https://www.cnblogs.com/onionQAQ/p/10738436.html

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