HYSBZ4154-Generating Synergy

[Ipsc2015]Generating Synergy

Time Limit: 10 Sec  Memory Limit: 512 MB
Submit: 712  Solved: 277
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Description

给定一棵以1为根的有根树,初始所有节点颜色为1,每次将距离节点a不超过l的a的子节点染成c,或询问点a的颜色

Input

第一行一个数T,表示数据组数

接下来每组数据的第一行三个数n,c,q表示结点个数,颜色数和操作数

接下来一行n-1个数描述2..n的父节点

接下来q行每行三个数a,l,c

若c为0,表示询问a的颜色

否则将距离a不超过l的a的子节点染成c

Output

设当前是第i个操作,y_i为本次询问的答案(若本次操作是一个修改则y_i为0),令z_i=i*y_i,请输出z_1+z_2+...+z_q模10^9+7

Sample Input

1
4 3 7
1 2 2
3 0 0
2 1 3
3 0 0
1 0 2
2 0 0
4 1 1
4 0 0

Sample Output

32

HINT

第1,3,5,7的询问的答案分别为1,3,3,1,所以答案为 1*1+2*0+3*3+4*0+5*3+6*0+7*1=32.

数据范围:

对于100%的数据T<=6,n,m,c<=10^5,

1<=a<=n,0<=l<=n,0<=c<=c

Source

解题思路：kd-tree，可以以每个节点的 DFS 序作为横坐标，深度作为纵坐标，那么就可以把一棵树放在坐标系里了，一次 a 节点，距离为 l 的染色就是纵坐标从 dep[a] 到 dep[a] + l，横坐标从 dfsx[a] 到 dfsx[a] + size[a] - 1 的矩形，dep[a] 表示节点 a 的深度，dfsx[a] 指节点 a 的 DFS 序，size[a] 表示以节点 a 为根的子树大小。更新需要标记下传像线段树一样弄一个标记不断向下移即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cctype>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const LL mod = 1e9 + 7;const int N = 200003;const int demension = 2;//二维struct node{int pos[demension];int ma[demension], mi[demension];int l, r, val, lazy;}a[N], x;int cmpDem;//以第cmpDem维作比较int root, n, c, q, cnt, u, l, L[N], R[N], deep[N];int s[N], nt[N], e[N];void dfs(int k,int dep){L[k] = cnt++, deep[k] = dep;for (int i = s[k]; ~i; i = nt[i]) dfs(e[i], dep + 1);R[k] = cnt - 1;}bool cmp(const node &a, const node&b){return a.pos[cmpDem] < b.pos[cmpDem];}void Merge(int k){for (int i = 0; i < demension; i++){if (a[k].l){a[k].ma[i] = max(a[k].ma[i], a[a[k].l].ma[i]);a[k].mi[i] = min(a[k].mi[i], a[a[k].l].mi[i]);}if (a[k].r){a[k].ma[i] = max(a[k].ma[i], a[a[k].r].ma[i]);a[k].mi[i] = min(a[k].mi[i], a[a[k].r].mi[i]);}}}int build(int l, int r, int k){if (l > r) return 0;int mid = (l + r) / 2;//以第mid个元素为中心排序cmpDem = k;nth_element(a + l, a + mid, a + r + 1, cmp);//左右子树a[mid].l = build(l, mid - 1, (k + 1) % demension);a[mid].r = build(mid + 1, r, (k + 1) % demension);Merge(mid);return mid;}int check(int k){if (x.mi[0] <= a[k].mi[0] && a[k].ma[0] <= x.ma[0] && x.mi[1] <= a[k].mi[1] && a[k].ma[1] <= x.ma[1]) return 1;return 0;}int check1(int k){if (x.pos[0] <= a[k].ma[0] && a[k].mi[0] <= x.pos[0] && x.pos[1] <= a[k].ma[1] && a[k].mi[1] <= x.pos[1]) return 1;return 0;}int check2(int k){int ans = 0;for (int i = 0; i < demension; i++){int temp = 1;if (x.ma[i] < a[k].mi[i] || x.mi[i] > a[k].ma[i]) temp = 0;ans += temp;}if (ans == 2) return 1;return 0;}int check3(int k){if (x.mi[0] <= a[k].pos[0] && x.ma[0] >= a[k].pos[0] && x.mi[1] <= a[k].pos[1] && x.ma[1] >= a[k].pos[1]) return 1;return 0;}void update(int k){if (check(k)) { a[k].lazy = a[k].val = x.val; return; }if (a[k].lazy) {a[a[k].l].lazy = a[a[k].r].lazy = a[k].val = a[k].lazy, a[k].lazy = 0;}if (check3(k)) a[k].val = x.val;if (a[k].l&&check2(a[k].l)) update(a[k].l);if (a[k].r&&check2(a[k].r)) update(a[k].r);}int query(int k){if (a[k].lazy) { a[a[k].l].lazy = a[a[k].r].lazy = a[k].val = a[k].lazy, a[k].lazy = 0; }if (x.pos[0] == a[k].pos[0]) return a[k].val;if (!a[k].l && !a[k].r) return -1;int ans = -1;if (a[k].l && check1(a[k].l))ans = query(a[k].l);if (ans != -1) return ans;return query(a[k].r);}int main(){int t;scanf("%d", &t);while (t--){scanf("%d%d%d", &n, &c, &q);cnt = 0;memset(s, -1, sizeof s);for (int i = 2; i <= n; i++){scanf("%d", &u);nt[cnt] = s[u], s[u] = cnt, e[cnt++] = i;}dfs(cnt = 1, 1);for (int i = 1; i <= n; i++){a[i].pos[0] = L[i], a[i].pos[1] = deep[i], a[i].val = 1, a[i].lazy = 0;for (int j = 0; j < 2; j++) a[i].ma[j] = a[i].mi[j] = a[i].pos[j];}LL ans = 0;root = build(1, n, 0);for(int i = 1; i <= q; i++){scanf("%d%d%d", &u, &l, &c);if (!c){x.pos[0] = L[u], x.pos[1] = deep[u];int temp = query(root);ans = (ans + 1LL * temp * i % mod) % mod;}else{x.mi[0] = L[u], x.ma[0] = R[u];x.mi[1] = deep[u], x.ma[1] = deep[u] + l, x.val = c;update(root);}}printf("%lld\n", ans);}return 0;}