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#include #include #include 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 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 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("\n"); */ if(f[1] >= inf) puts("-1"); else printf("%lld\n", f[1]); return 0;}
感觉还是超过了能力范围……