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  • 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;
    	 
    }
    
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  • 原文地址:https://www.cnblogs.com/bingoyes/p/11175974.html
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