HDU - 5770 Treasure 思维 + 扫描线 + 线段树 (看题解)
没想出来, 感觉不应该啊, 没有想到转换成二维上的点的问题。
对于对钥匙和宝藏(u, v), 如果lca != u && lca != v 那么起点从u子树出发, 终点在v子树就能得到贡献。
子树在dfs序下是连续一段, 所以就对应到二维平面一个矩形加上一个数值, 求值最大的点。
对于lca == u || lca == v同样可以讨论出来。
还有一种情况就是u == v, 我们先把贡献都加上, 然后对于不经过u 的所有路径进去这个贡献。
然后扫描线扫一遍就好了。
#pragma GCC optimize(2) #pragma GCC optimize(3) #include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = (int)1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} const int LOG = 17; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 struct SegmentTree { int mx[N << 2], lazy[N << 2]; void build(int l, int r, int rt) { mx[rt] = lazy[rt] = 0; if(l == r) return; int mid = l + r >> 1; build(lson); build(rson); } inline void push(int rt) { if(lazy[rt]) { mx[rt << 1] += lazy[rt]; mx[rt << 1 | 1] += lazy[rt]; lazy[rt << 1] += lazy[rt]; lazy[rt << 1 | 1] += lazy[rt]; lazy[rt] = 0; } } void update(int L, int R, int val, int l, int r, int rt) { if(R < l || r < L || R < L) return; if(L <= l && r <= R) { mx[rt] += val; lazy[rt] += val; return; } push(rt); int mid = l + r >> 1; update(L, R, val, lson); update(L, R, val, rson); mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]); } inline int getMax() { return mx[1]; } } Tree; int n, m; int sig_val; int add_val[N]; int in[N], ot[N], idx; int depth[N], pa[N][LOG]; vector<int> G[N]; void dfs(int u, int fa) { depth[u] = depth[fa] + 1; pa[u][0] = fa; in[u] = ++idx; for(int i = 1; i < LOG; i++) { pa[u][i] = pa[pa[u][i - 1]][i - 1]; } for(auto &v : G[u]) { if(v == fa) continue; dfs(v, u); } ot[u] = idx; } int getLca(int u, int v) { if(depth[u] < depth[v]) swap(u, v); int d = depth[u] - depth[v]; for(int i = LOG - 1; i >= 0; i--) { if(d >> i & 1) { u = pa[u][i]; } } if(u == v) return u; for(int i = LOG - 1; i >= 0; i--) { if(pa[u][i] != pa[v][i]) { u = pa[u][i]; v = pa[v][i]; } } return pa[u][0]; } inline int go(int u, int step) { for(int i = LOG - 1; i >= 0; i--) { if(step >> i & 1) { u = pa[u][i]; } } return u; } int L_cnt; struct Line { int x, y1, y2, val; bool operator < (const Line &rhs) const { return x < rhs.x; } } L[N * 10]; void init() { idx = L_cnt = sig_val = 0; for(int i = 1; i <= n; i++) { add_val[i] = 0; G[i].clear(); } Tree.build(1, n, 1); } int main() { int cas = 0; int T; scanf("%d", &T); while(T--) { scanf("%d%d", &n, &m); init(); for(int i = 1; i < n; i++) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } dfs(1, 0); for(int i = 1; i <= m; i++) { int u, v, w; scanf("%d%d%d", &u, &v, &w); int lca = getLca(u, v); if(u == v) { sig_val += w; add_val[u] -= w; } else { if(u == lca) { u = go(v, depth[v] - depth[u] - 1); if(in[u] > 1) { L[++L_cnt] = Line{1, in[v], ot[v], w}; L[++L_cnt] = Line{in[u], in[v], ot[v], -w}; } if(ot[u] < n) { L[++L_cnt] = Line{ot[u] + 1, in[v], ot[v], w}; L[++L_cnt] = Line{n + 1, in[v], ot[v], -w}; } } else if(v == lca) { v = go(u, depth[u] - depth[v] - 1); if(in[v] > 1) { L[++L_cnt] = Line{in[u], 1, in[v] - 1, w}; L[++L_cnt] = Line{ot[u] + 1, 1, in[v] - 1, -w}; } if(ot[v] < n) { L[++L_cnt] = Line{in[u], ot[v] + 1, n, w}; L[++L_cnt] = Line{ot[u] + 1, ot[v] + 1, n, -w}; } } else { L[++L_cnt] = Line{in[u], in[v], ot[v], w}; L[++L_cnt] = Line{ot[u] + 1, in[v], ot[v], -w}; } } } for(int u = 1; u <= n; u++) { if(add_val[u]) { for(auto &v : G[u]) { if(v == pa[u][0]) continue; L[++L_cnt] = Line{in[v], in[v], ot[v], add_val[u]}; L[++L_cnt] = Line{ot[v] + 1, in[v], ot[v], -add_val[u]}; } if(pa[u][0]) { int v = u; L[++L_cnt] = Line{1, 1, in[v] - 1, add_val[u]}; L[++L_cnt] = Line{1, ot[v] + 1, n, add_val[u]}; L[++L_cnt] = Line{in[v], 1, in[v] - 1, -add_val[u]}; L[++L_cnt] = Line{in[v], ot[v] + 1, n, -add_val[u]}; L[++L_cnt] = Line{ot[v] + 1, 1, in[v] - 1, add_val[u]}; L[++L_cnt] = Line{ot[v] + 1, ot[v] + 1, n, add_val[u]}; } } } int ans = -inf; sort(L + 1, L + 1 + L_cnt); for(int i = 1, j = 1; i <= n; i++) { while(j <= L_cnt && L[j].x <= i) { if(L[j].y1 <= L[j].y2) { Tree.update(L[j].y1, L[j].y2, L[j].val, 1, n, 1); } j++; } chkmax(ans, sig_val + Tree.getMax()); } printf("Case #%d: ", ++cas); printf("%d\n", ans); } return 0; } /* */