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

#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("\n\n-------------------\n\n")

#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];
inline void add(int u,int v,long long w){
}

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){
}
}

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\n", MST + sum);

return 0;

}

posted @ 2019-07-12 14:42  邱涵的秘密基地  阅读(74)  评论(0编辑  收藏  举报