CF671D Roads in Yusland

做法一：   

线段树合并.     

令 $\mathrm{dp[x][i]}$ 表示以 $\mathrm{x}$ 节点为根的子树全部被覆盖且延伸到了深度为 $\mathrm{i}$ 的祖先.  

考虑 $\mathrm{x},\mathrm{y}$ 两个子树如何合并：    

有 $\mathrm{dp'[x][i]=dp[x][i]+}$ $\mathrm{min}${$\mathrm{dp[y]}$}, $\mathrm{dp[y]}$ 同理.     

这个可以用线段树来整体维护，每次就是将 $\mathrm{x}$ 与 $\mathrm{y}$ 子树进行区间加法.     

更新以 $\mathrm{x}$ 点为终点的链是一个单点更新，也可以用线段树来维护.    

两个子树合并用线段树合并即可, 合并区间时将区间最小值取 $\min$.  

#include <cstdio>
#include <cstring> 
#include <vector> 
#include <algorithm>   
#define N  300009 
#define ll long long 
#define pb push_back 
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std;   
const ll in2 = (ll)1e16;   
const ll inf = (ll)1e17;  
int n,m,tot, rt[N], dep[N];                
vector<int>G[N];      
struct Line { 
    int x, c; 
    Line(int x=0,int c=0):x(x),c(c){}  
};  
vector<Line>q[N];                   
struct data {
    ll mi;  
    int ls, rs;         
    data() { mi = inf, ls = rs = 0; }      
}s[N * 30];          
ll add[N * 30]; 
void mark(int x, ll v) {
    if(!x) return ;                   
    add[x] = min(inf, add[x] + v);     
    s[x].mi = min(inf, s[x].mi + v);   
}
void pushdown(int x) {
    if(add[x]) {
        mark(s[x].ls, add[x]);  
        mark(s[x].rs, add[x]);  
    } 
    add[x] = 0; 
}
void pushup(int x) {
    s[x].mi = min(s[s[x].ls].mi, s[s[x].rs].mi);   
}
void update(int &x, int l, int r, int p, ll v) {               
    if(!x) x = ++ tot, s[x].mi = inf; 
    s[x].mi = min(s[x].mi, v);        
    if(l == r) return ; 
    pushdown(x);  
    int mid = ( l + r ) >> 1;         
    if(p <= mid) {
        update(s[x].ls, l, mid, p, v); 
    }
    else {
        update(s[x].rs, mid + 1, r, p, v); 
    }
    pushup(x); 
}        
int merge(int l, int r, int x, int y) {
    if(!x || !y) return x + y;  
    s[x].mi = min(s[x].mi, s[y].mi);      
    if(l == r) return x;   
    int mid = (l + r) >> 1;             
    pushdown(x);  
    pushdown(y);  
    s[x].ls = merge(l, mid, s[x].ls, s[y].ls);  
    s[x].rs = merge(mid + 1, r, s[x].rs, s[y].rs); 
    return x;   
}       
ll query(int l, int r, int x, int L, int R) {
    if(!x || l > r || L > R) return inf;  
    if(l >= L && r <= R) return s[x].mi;   
    pushdown(x);  
    int mid = (l + r) >> 1;  
    if(L <= mid && R > mid) 
        return min(query(l, mid, s[x].ls, L, R), query(mid + 1, r, s[x].rs, L, R)); 
    else if(L <= mid) 
        return query(l, mid, s[x].ls, L, R); 
    else return query(mid + 1, r, s[x].rs, L, R);   
}
void dfs(int x, int ff) {
    dep[x] = dep[ff] + 1;   
    int son = 0; 
    for(int i=0;i<G[x].size();++i) {  
        int y=G[x][i]; 
        if(y==ff) continue;  
        dfs(y, x);  
        if(!son) { ++son, rt[x]=rt[y]; continue; }       

        ++ son ;   
        // 合并 v    
        ll px = query(1, n, rt[x], 1, dep[x]); 
        ll py = query(1, n, rt[y], 1, dep[x]);       
        mark(rt[x], py); 
        mark(rt[y], px);                            
        rt[x] = merge(1, n, rt[x], rt[y]);       
    }
    if(son == 0) {
        update(rt[x], 1, n, dep[x], 0ll);     
    }
    // 合并完了 x 
    for(int i = 0; i < q[x].size() ; ++ i) {      
        int tp = q[x][i].x;  
        ll v = (ll)q[x][i].c; 
        ll px = query(1, n, rt[x], 1, dep[x]);   
        if(v + px < inf) {
            update(rt[x],1, n, dep[tp], v + px);             
        }
    }                            
}       
void ini(int x, int ff) {
    dep[x] = dep[ff] + 1; 
    for(int i = 0 ; i < G[x].size(); ++ i) 
        if(G[x][i] != ff) ini(G[x][i], x);  
}
int main() {
    // setIO("input");    
    s[0].mi = inf ;  
    scanf("%d%d",&n,&m);  
    for(int i=1;i<n;++i) {
        int x, y; 
        scanf("%d%d",&x,&y); 
        G[x].pb(y); 
        G[y].pb(x);  
    }
    ini(1, 0); 
    for(int i=1;i<=m;++i) {
        int x,y,z; 
        scanf("%d%d%d",&x,&y,&z);  
        if(x == y) continue;  
        if(dep[y] > dep[x]) swap(x, y);    
        q[x].pb(Line(y, z));    
    }           
    dfs(1, 0); 
    ll ans = query(1, n, rt[1], 1, 1);             
    if(ans > in2) printf("-1\n"); 
    else printf("%lld\n", ans);   
    return 0;
}