P4556 [Vani有约会]雨天的尾巴 /【模板】线段树合并 （树上差分+线段树合并）

显然的树上差分问题，最后要我们求每个点数量最多的物品，考虑对每个点建议线段树，查询子树时将线段树合并可以得到答案。

用动态开点的方式建立线段树，注意离散化。

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 const int N = 1e5 + 10;
  4 struct node {
  5     int lc, rc, dat, pos;//dat记录最多的物品的次数，pos记录位置 
  6 }tr[N * 20 * 4];
  7 int head[N], to[N << 1], nxt[N << 1], tot;
  8 int n, m, num, cnt, t, ans[N];
  9 int d[N], st[N][20], rt[N], X[N], Y[N], Z[N], val[N];
 10 inline int read() {
 11     int x = 0,f = 1;char ch = getchar();
 12     while (ch<'0' || ch>'9') { if(ch == '-') f = -1;ch = getchar(); }
 13     while (ch >= '0'&&ch <= '9') x = x * 10 + ch - '0',ch = getchar();
 14     return x * f;
 15 }
 16 void add(int x, int y) {
 17     nxt[++tot] = head[x];
 18     head[x] = tot;
 19     to[tot] = y;
 20 }
 21 
 22 void dfs(int u, int f) {
 23      for (int i = head[u]; i; i = nxt[i]) {
 24          int v = to[i];
 25          if (d[v]) continue;
 26          d[v] = d[u] + 1;
 27          st[v][0] = u;
 28          for (int j = 1; j <= t; j++)
 29              st[v][j] = st[st[v][j-1]][j-1];
 30          dfs(v, u);
 31      }
 32 }
 33 
 34 int lca(int x, int y) {
 35     if (d[x] > d[y]) swap(x, y);
 36     for (int i = t; i >= 0; i--) 
 37         if (d[st[y][i]] >= d[x]) y = st[y][i];
 38     if (x == y) return x;
 39     for (int i = t; i >= 0; i--)
 40         if (st[x][i] != st[y][i]) x = st[x][i], y = st[y][i];
 41     return st[x][0]; 
 42 }
 43 
 44 void insert(int p, int l, int r, int val, int k) {
 45     if (l == r) {
 46         tr[p].dat += k;
 47         tr[p].pos = tr[p].dat ? l : 0;
 48         return ;
 49     }
 50     int mid = (l + r) >> 1;
 51     if(val <= mid) {
 52         if (!tr[p].lc)    tr[p].lc = ++num;//动态开点 
 53         insert(tr[p].lc, l, mid, val, k);
 54     }
 55     else {
 56         if (!tr[p].rc) tr[p].rc = ++num;
 57         insert(tr[p].rc, mid + 1, r, val, k);
 58     }
 59     tr[p].dat = max(tr[tr[p].lc].dat, tr[tr[p].rc].dat);
 60     tr[p].pos = tr[tr[p].lc].dat >= tr[tr[p].rc].dat ? tr[tr[p].lc].pos : tr[tr[p].rc].pos;
 61 }
 62 
 63 int merge(int p, int q, int l, int r) {//线段树合并 
 64     if (!p || !q) return p + q;
 65     if (l == r) {
 66         tr[p].dat += tr[q].dat;
 67         tr[p].pos = tr[p].dat ? l : 0;
 68         return p;
 69     }
 70     int mid = (l + r) >> 1;
 71     tr[p].lc = merge(tr[p].lc, tr[q].lc, l, mid);
 72     tr[p].rc = merge(tr[p].rc, tr[q].rc, mid + 1, r);
 73     tr[p].dat = max(tr[tr[p].lc].dat, tr[tr[p].rc].dat);
 74     tr[p].pos = tr[tr[p].lc].dat >= tr[tr[p].rc].dat ? tr[tr[p].lc].pos : tr[tr[p].rc].pos;
 75     return p;
 76 }
 77 
 78 void solve(int x) {
 79     for (int i = head[x]; i; i = nxt[i]) {
 80         int y = to[i];
 81         if (d[y] <= d[x]) continue;
 82         solve(y);
 83         rt[x] = merge(rt[x], rt[y], 1, cnt);
 84     }
 85     ans[x] = tr[rt[x]].pos;
 86 }
 87 
 88 int main() {
 89     n = read(); m = read();
 90     t = log2(n) + 1;
 91     for (int i = 1; i < n; i++) {
 92         int x = read(), y = read();
 93         add(x, y), add(y, x);
 94     }
 95     d[1] = 1, dfs(1,0);
 96     for (int i = 1; i <= n; i++) rt[i] = ++num;
 97     for (int i = 1; i <= m; i++) {
 98         X[i] = read(); Y[i] = read(); Z[i] = read();
 99         val[i] = Z[i];
100     }
101     sort(val + 1, val + m + 1);//离散化 
102     cnt = unique(val + 1, val + m + 1) - val - 1;
103     for (int i = 1; i <= m; i++) {
104         int x = X[i], y = Y[i];
105         int z = lower_bound(val + 1, val + cnt + 1, Z[i]) - val;
106         int p = lca(x, y);
107         //树上差分 
108         insert(rt[x], 1, cnt, z, 1);
109         insert(rt[y], 1, cnt, z, 1);
110         insert(rt[p], 1, cnt, z, -1);
111         if (st[p][0]) insert(rt[st[p][0]], 1, cnt, z, -1);
112     }
113     solve(1);
114     for (int i = 1; i <= n; i++) printf("%d\n", val[ans[i]]);
115     return 0;
116 }

标签：lc,int,P4556,线段,合并,pos,tr,dat,rc
来源： https://www.cnblogs.com/yhxnoerror/p/16514516.html