CF 671D Roads in Yusland

弄完之后点进去一看，竟然是div1的D题……最近真是天天被题虐哭

推荐这一篇博客 https://www.cnblogs.com/Sakits/p/8085598.html 感觉讲清楚了，也是基本照着这个写的

一开始题意没有读清楚，这题保证了所以树链都是从下往上的，所以才可以设计dp。

要把询问映射到树的dfs序上，快速计算子树中的最小值，想到可以使用线段树。

这个线段树还真是不好写。

Code：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N = 5e5 + 5;
const ll inf = (ll)1 << 60;

int n, m, tot = 0, head[N], dfsc = 0, st[N], ed[N];
ll f[N];
vector <int> in[N], out[N];

struct Edge {
    int to, nxt;
} e[N << 1];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

struct Querys {
    int x, y, pos;
    ll cost;
} q[N];

bool cmp(const Querys a, const Querys b) {
    return a.pos < b.pos;
}

template <typename T>
inline void read(T &X) {
    X = 0;
    char ch = 0;
    T op = 1;
    for(; ch > '9'|| ch < '0'; ch = getchar())
        if(ch == '-') op = -1;
    for(; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

void dfs(int x, int fat) {
    st[x] = ++dfsc;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        dfs(y, x);
    }
    ed[x] = dfsc;
}

inline void chkMin(ll &x, ll y) {
    if(y < x) x = y;
}

inline ll min(ll x, ll y) {
    return x > y ? y : x;
}

namespace SegT {
    ll s[N << 2], tag[N << 2];
    
    #define lc p << 1
    #define rc p << 1 | 1
    #define mid ((l + r) >> 1)
    
    inline void up(int p) {
        if(p) s[p] = min(s[lc], s[rc]);
    }
    
    inline void done(int p, ll v) {
        tag[p] = min(inf, tag[p] + v);
        s[p] = min(inf, s[p] + v);
    }
    
    inline void down(int p) {
        if(tag[p] == 0LL) return;
        done(lc, tag[p]), done(rc, tag[p]);
        tag[p] = 0LL;
    }
    
    void build(int p, int l, int r) {
        s[p] = inf, tag[p] = 0LL;
        if(l == r) return;
        
        build(lc, l, mid);
        build(rc, mid + 1, r);
    }
    
    void modifyP(int p, int l, int r, int x, ll v) {
        if(x == l && r == x) {
            s[p] = v;
            return;
        }
        
        down(p);
        if(x <= mid) modifyP(lc, l, mid, x, v);
        else modifyP(rc, mid + 1, r, x, v);
        up(p);
    }
    
    void modify(int p, int l, int r, int x, int y, ll v) {
        if(x > y) return;
        if(x <= l && y >= r) {
            done(p, v);
            return;
        }
        
        down(p);
        if(x <= mid) modify(lc, l, mid, x, y, v);
        if(y > mid) modify(rc, mid + 1, r, x, y, v);
        up(p);
    }
    
    ll query(int p, int l, int r, int x, int y) {
        if(x > y) return inf;
        if(x <= l && y >= r) return s[p];
        down(p);
        ll res = inf;
        if(x <= mid) res = min(res, query(lc, l, mid, x, y));
        if(y > mid) res = min(res, query(rc, mid + 1, r, x, y));
        return res;
    }
    
    #undef mid
    #undef lc
    #undef rc
    
} using namespace SegT;

inline int bfind(int x) {
    ll ln = 1, rn = m + 1, mid, res;
    for(; ln <= rn; ) {
        mid = (ln + rn) / 2;
        if(q[mid].pos >= x) res = mid, rn = mid - 1;
        else ln = mid + 1;
    }
    return res;
}   

void solve(int x, int fat) {
    ll sum = 0;
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        solve(y, x);
        sum = min(inf, sum + f[y]);
    }
    
    if(x == 1) {
        f[1] = sum;
        return;
    }
    
    for(unsigned int i = 0; i < in[x].size(); i++)
        modifyP(1, 1, m, in[x][i], min(inf, q[in[x][i]].cost + sum));
    for(unsigned int i = 0; i < out[x].size(); i++)
        modifyP(1, 1, m, out[x][i], inf);
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(y == fat) continue;
        modify(1, 1, m, bfind(st[y]), bfind(ed[y] + 1) - 1, sum - f[y]);
    }
    
    f[x] = query(1, 1, m, bfind(st[x]), bfind(ed[x] + 1) - 1);
}

int main() {
    read(n), read(m);
    build(1, 1, m);
    for(int x, y, i = 1; i < n; i++) {
        read(x), read(y);
        add(x, y), add(y, x);
    }
    dfs(1, 0);
    
    for(int i = 1; i <= m; i++) {
        read(q[i].x), read(q[i].y), read(q[i].cost);
        q[i].pos = st[q[i].x];
    }
    
    sort(q + 1, q + 1 + m, cmp);
    q[m + 1].pos = n + 1;
    
    for(int i = 1; i <= m; i++) 
        in[q[i].x].push_back(i), out[q[i].y].push_back(i);
    
    solve(1, 0);
    
/*    for(int i = 1; i <= n; i++)
        printf("%lld ", f[i]);
    printf("
");   */
    
    if(f[1] >= inf) puts("-1");
    else printf("%lld
", f[1]);
    
    return 0;
}

感觉还是超过了能力范围……