BZOJ1977/LuoguP4180【模板】严格次小生成树[BJWC2010] （次小生成树）

这道题本身思维难度不大，但综合性强，细节多
在其上浪一个早上，你的
最小生成树 树链剖分 线段树 DEBUG能力...
都大幅提升
细节与思路都在代码里面了。
欢迎hack.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define ll long long
#define Fill(a,b) memset(a, b, sizeof(a))

//#define ON_DUBUG

#ifdef ON_DUBUG

#define D_e_Line printf("

-------------------

")

#else

#define D_e_Line ;

#endif

struct ios{
	template<typename ATP>ios& operator >> (ATP &x){
		x = 0; int f = 1; char c;
		for(c = getchar(); c < '0' || c > '9'; c=getchar()) if(c == '-') f = -1;
		while(c >= '0' && c <='9') x = x * 10 + (c ^ '0'), c = getchar();
		x *= f;
		return *this;
	}
}io;

using namespace std;

//#define TEST_INPUT
#ifdef TEST_INPUT
int main(){
	while(1){
		ll x;
		io >> x;
		cout << x << endl;
	}
}
#endif

const int N = 100007;
const int M = 300007;

int n, m;

struct Node{
	int x,y,w,tag;
	bool operator < (const Node &b)const{
		return w < b.w;
	}
}a[M];
struct Edge{
	int nxt,pre;
	long long w;	
}e[M<<1];
int head[N],cntEdge;
inline void add(int u,int v,long long w){
	e[++cntEdge] = (Edge){head[u], v, w}, head[u] = cntEdge;
}

long long MST;
int f[N];
inline int Find(int x){
	return x == f[x] ? x : f[x] = Find(f[x]);
}
inline void Kruskal(){
	R(i,1,n) f[i] = i;//Always remember, Fool orphan! Get your father at the first time!
	sort(a + 1, a + m + 1);
	int tot = 1;
	R(i,1,m){
		int p = Find(a[i].x), q =Find(a[i].y);
		if(p != q){
			f[p] = q;
			MST += a[i].w;
			a[i].tag = true;
			if(++tot >= n)
				return;
		}
	}
	
}

int wSon[N];
int fa[N],son[N],siz[N],dep[N];
inline void DFS_First(int u,int father){
	dep[u] = dep[father] + 1, siz[u] = 1, fa [u] = father;
	for(register int i = head[u]; i; i = e[i].nxt){
		int v = e[i].pre;
		if(v == father) continue;
		DFS_First(v, u);
		siz[u] += siz[v];
		if(!son[u] || siz[v] > siz[son[u]]){
			son[u] = v;
			//Treat value(edge) as value(node)
			wSon[u] = e[i].w;
		}
	}
}
int top[N],rnk[N],dfn[N],dfnIndex;
inline void DFS_Second(int u,int ancester,int w){
	top[u] = ancester, dfn[u] = ++dfnIndex, rnk[dfnIndex] = w;
	if(!son[u]) return;
	DFS_Second(son[u], ancester, wSon[u]);//e[i].w or wSon[u] ? // The answer is wSon[u] because it's the weight of this node
	for(register int i = head[u]; i; i = e[i].nxt){
		int v = e[i].pre;
		if(v != fa[u] && v != son[u])
			DFS_Second(v, v, e[i].w);//e[i].w or wSon[u] ? // The answer is e[i].w because it's the cost to the next node.
	}
}

#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
struct SegmentTree{
	int mx,sc;
}t[M<<2];
inline void Pushup(int rt){
	t[rt].mx = Max(t[rt<<1].mx, t[rt<<1|1].mx);
	t[rt].sc = Max(t[rt].mx != t[rt<<1].mx ? t[rt<<1].mx : t[rt<<1].sc, t[rt].mx != t[rt<<1|1].mx ? t[rt<<1|1].mx : t[rt<<1|1].sc);
}
inline void Build(int rt,int l,int r){
	if(l == r){
		t[rt].mx = rnk[l];
		t[rt].sc = -1;
		//Pushup(rt);//Do there really need pushup? // No! You do not have to pull a tree you just put into the groud.
		return;
	}
	int mid = l + r >> 1;
	Build(lson), Build(rson);
	Pushup(rt);
}
SegmentTree Query(int rt,int l,int r,int L,int R){
	if(L <= l && r <= R) return t[rt];
	
	SegmentTree tmp_1, tmp_2, res;
	tmp_1.mx = tmp_1.sc = tmp_2.mx = tmp_2.sc;//Need initial? And why?
	/*
	There is a strange problem
		if I don't initial, the program will give a wrong answer.
		But if I write:
			tmp_1.mx = tmp_1.sc = tmp_2.mx = tmp_2.sc
		It'll be right.
		Pay attention, I do not write:
			tmp_1.mx = tmp_1.sc = tmp_2.mx = tmp_2.sc = 0
		
		So what happend to this initial sentence?
			I believe you can find out by yourself.
	*/

	int mid = l + r >> 1;
	//D_e_Line;
	if(L <= mid) tmp_1 = Query(lson, L, R);
	if(mid < R) tmp_2 = Query(rson, L, R);

	res.mx = Max(tmp_1.mx, tmp_2.mx);
	res.sc = Max(res.mx != tmp_1.mx ? tmp_1.mx : tmp_1.sc, res.mx != tmp_2.mx ? tmp_2.mx : tmp_2.sc);
	
	return res;
}

inline long long QueryPath(int u,int v,long long w){
	long long sum = 0;
	while(top[u] != top[v]){
		if(dep[top[u]] < dep[top[v]]) u^=v^=u^=v;
		SegmentTree tmp = Query(1, 1, n, dfn[top[u]], dfn[u]);
		sum = Max(sum, tmp.mx != w ? tmp.mx : tmp.sc);
		u = fa[top[u]];
	}
	if(u == v) return sum;//Is this so important? Is it the criminal of the endless of my program?//Well, for the second question, it may not. //And if I ignore it, it doesn't matter, but as it can make my program runs faster, why not?
	if(dep[u] < dep[v]) u^=v^=u^=v;
	SegmentTree tmp = Query(1, 1, n, dfn[v]+1, dfn[u]);//I'm not sure when I need use 'dfn[v]+1' 
	sum = Max(sum, tmp.mx != w ? tmp.mx : tmp.sc);
	return sum;
}

int main(){
	io >> n >> m;
	R(i,1,m){
		int u,v,w;
		io >> u >> v >> w;
		a[i] = (Node){u, v, w, 0};
	}
	
	Kruskal();
	 
	R(i,1,m){//Here ought to be m, not n! It's going through all the edges
		if(a[i].tag == true){
			add(a[i].x, a[i].y, a[i].w);
			add(a[i].y, a[i].x, a[i].w);
		}
	}
	
	DFS_First(1, 0);
	DFS_Second(1, 0, 0);
	
	Build(1, 1, n);
	
	int sum = 2147483647;
	R(i,1,m){
		if(a[i].tag == false){
			sum = Min(sum, a[i].w - QueryPath(a[i].x, a[i].y, a[i].w));
		}
	}
	
	printf("%lld
", MST + sum);
	
	return 0;
	 
}