BZOJ1977: [BeiJing2010组队]次小生成树 Tree

1977: [BeiJing2010组队]次小生成树 Tree

题意:求严格次小生成树


我为什么要单独发这篇呢

因为愚蠢的我不停换写法最后发现是因为没开long long所以wa掉的

很简单,次小生成树是由mst换一条边得到的

就是枚举非树边,加入后会形成一个环,求环上的最大值和严格次大值与这条非树边换一换

用倍增可以做到O(mn)-->O(mlogn)

我写了两种求lca时同时求路径上最大值/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;
}

Copyright:http://www.cnblogs.com/candy99/
posted @ 2018-07-06 13:07  Candy?  阅读(270)  评论(0编辑  收藏  举报