BZOJ1977: [BeiJing2010組隊]次小生成樹 Tree
阿新 • • 發佈:2018-07-06
++ spa ring mes code ref namespace urn sync
1977: [BeiJing2010組隊]次小生成樹 Tree
題意:求嚴格次小生成樹
我為什麽要單獨發這篇呢
因為愚蠢的我不停換寫法最後發現是因為沒開long long所以wa掉的
很簡單,次小生成樹是由mst換一條邊得到的
就是枚舉非樹邊,加入後會形成一個環,求環上的最大值和嚴格次大值與這條非樹邊換一換
用倍增可以做到O(mn)-->O(mlogn)
我寫了兩種求lca時同時求路徑上最大值/次大值的做法
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <set> using namespace std; typedef long long ll; const int N = 1e5+5, M = 3e5+5; int n, m; struct edge {int v, ne, w;} e[M<<2]; int cnt, h[N]; inline void ins(int u, int v, int w) { e[++cnt] = (edge) {v, h[u], w}; h[u] = cnt; e[++cnt] = (edge) {u, h[v], w}; h[v] = cnt; } struct meow { int u, v, w; bool operator < (const meow &r) const { return w < r.w; } } a[M]; bool mark[M]; namespace mst { int fa[N]; int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);} ll kruskal() { for(int i=1; i<=n; i++) fa[i] = i; sort(a+1, a+1+m); int cnt = 0; ll ans = 0; for(int i=1; i<=m; i++) { int x = find(a[i].u), y = find(a[i].v); if(x != y) { //printf("use (%d, %d) %d\n", a[i].u, a[i].v, a[i].w); fa[y] = x, ans += a[i].w, mark[i] = 1; ins(a[i].u, a[i].v, a[i].w); if(++cnt == n-1) break; } } return ans; } } int deep[N], fa[N][20], vis[N]; #define fir first #define sec second pair<int, int> val[N][20]; pair<int, int> merge(pair<int, int> a, pair<int, int> b) { pair<int, int> ans; if(a.fir == b.fir) ans.fir = a.fir, ans.sec = max(a.sec, b.sec); else if(a.fir > b.fir) ans.fir = a.fir, ans.sec = max(a.sec, b.fir); else ans.fir = b.fir, ans.sec = max(a.fir, b.sec); /* ans.fir = max(a.fir, b.fir); if(a.fir == b.fir) ans.sec = max(a.sec, b.sec); else { ans.sec = min(a.fir, b.fir); ans.sec = max(ans.sec, max(a.sec, b.sec)); } */ return ans; } void dfs(int u) { vis[u] = 1; for(int j=1; (1<<j)<=deep[u]; j++) { fa[u][j] = fa[fa[u][j-1]][j-1]; val[u][j] = merge(val[u][j-1], val[fa[u][j-1]][j-1]); } for(int i=h[u]; i; i=e[i].ne) { int v = e[i].v; if(vis[v]) continue; deep[v] = deep[u]+1; fa[v][0] = u; val[v][0] = make_pair(e[i].w, -1); dfs(v); } } pair<int, int> lca(int x, int y) { pair<int, int> ans(-1, -1); if(deep[x] < deep[y]) swap(x, y); int bin = deep[x] - deep[y]; for(int j=0; j<=17; j++) if((1<<j) & bin) ans = merge(ans, val[x][j]), x = fa[x][j]; if(x == y) return ans; for(int j=17; j>=0; j--) if(fa[x][j] != fa[y][j]) { ans = merge(ans, merge(val[x][j], val[y][j])); x = fa[x][j], y = fa[y][j]; } ans = merge(ans, merge(val[x][0], val[y][0])); return ans; } int lca2(int x, int y) { if(deep[x] < deep[y]) swap(x, y); int bin = deep[x] - deep[y]; for(int j=0; j<=17; j++) if((1<<j) & bin) x = fa[x][j]; if(x == y) return x; for(int j=17; j>=0; j--) if(fa[x][j] != fa[y][j]) x = fa[x][j], y = fa[y][j]; return fa[x][0]; } pair<int, int> cal(int x, int r) { pair<int, int> ans(-1e9, -1e9); int bin = deep[x] - deep[r]; for(int j=0; j<=17; j++) if((1<<j) & bin) ans = merge(ans, val[x][j]), x = fa[x][j]; return ans; } int main() { freopen("in", "r", stdin); ios::sync_with_stdio(false); cin.tie(); cout.tie(); cin >> n >> m; for(int i=1; i<=m; i++) { int u, v, w; cin >> u >> v >> w; a[i] = (meow) {u, v, w}; } ll mn = mst::kruskal(); dfs(1); int ans = 1e9+7; for(int i=1; i<=m; i++) if(!mark[i]) { int u = a[i].u, v = a[i].v, w = a[i].w; //pair<int, int> t = lca(u, v); //printf("hey (%d, %d) %d %d %d\n", u, v, w, t.fir, t.sec); int r = lca2(u, v); pair<int, int> t = merge(cal(u, r), cal(v, r)); if(w > t.fir) ans = min(ans, w - t.fir); if(w == t.fir) ans = min(ans, w - t.sec); } //printf("look %d %d\n", mn, ans); cout << mn + ans; }
BZOJ1977: [BeiJing2010組隊]次小生成樹 Tree