HDU 5044 - Tree （树链剖分）

Tree

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 567 Accepted Submission(s): 121

Problem Description

You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N

There are N - 1 edges numbered from 1 to N - 1.

Each node has a value and each edge has a value. The initial value is 0.

There are two kind of operation as follows:

● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.

● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.

After finished M operation on the tree, please output the value of each node and edge.

Input

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N, M ≤105),denoting the number of nodes and operations, respectively.

The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.

For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -105 ≤ k ≤ 105)

Output

For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.

The second line contains N integer which means the value of each node.

The third line contains N - 1 integer which means the value of each edge according to the input order.

Sample Input

2
4 2
1 2
2 3
2 4
ADD1 1 4 1
ADD2 3 4 2
4 2
1 2
2 3
1 4
ADD1 1 4 5
ADD2 3 2 4

Sample Output

Case #1:
1 1 0 1
0 2 2
Case #2:
5 0 0 5
0 4 0

Source

2014 ACM/ICPC Asia Regional Shanghai Online

Recommend

hujie | We have carefully selected several similar problems for you: 5053 5052 5051 5050 5049

Statistic | Submit | Discuss | Note

题意：

一棵树，初始点权和边权都为0。

两种操作

add1 u v d ： u -> v之间的所有点权值+d

add2 u v d ： u -> v之间的所有边权值+d

树链剖分入门模板题

比赛的时候才学的竟然很顺利的A了

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const double PI = (4.0*atan(1.0));

using namespace std;

const int maxn = 100000 + 20;
const int maxo = maxn * 4;

struct Edge {
int to,next;
} edge[maxn*2];

int head[maxn], tot;
int top[maxn]; // top[v]表示v所在的重链的顶端节点
int fa[maxn];  // 父亲节点
int deep[maxn]; // 深度
int num[maxn]; // num[v]表示以v为根的子树的节点数
int p[maxn]; // p[v]表示v与其父亲节点的连边在线段树中的位置
int fp[maxn]; // 和p数组相反
int son[maxn]; // 重儿子
int pos; // 对应区间计数器

void init() {
tot = 0;
memset(head, -1, sizeof(head));
pos = 1;
memset(son, -1, sizeof(son));
}

void addedge(int u, int v) {
edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;
}

// 第一遍dfs求出fa, deep, num, son
// u当前节点, pre为父节点, d为深度
void dfs1(int u, int pre, int d) {
deep[u] = d;
fa[u] = pre;
num[u] = 1;
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(v != pre) {
dfs1(v, u, d+1);
num[u] += num[v];
if(son[u] == -1 || num[v] > num[son[u]])
son[u] = v;
}
}
}

// 第二遍dfs求出top和p
// sp为当前点所在重链的顶节点
void dfs2(int u, int sp) {
top[u] = sp;
p[u] = pos++;
fp[p[u]] = u;
if(son[u] == -1) return;
dfs2(son[u], sp);
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(v != son[u] && v != fa[u])
dfs2(v, v);
}
}

// 区间+d，最后求每个点的值
// 离线标记法

LL Seg[maxn][2];

void SegAdd(int L, int R, LL v, int d) {
Seg[L][d] += v;
Seg[R+1][d] -= v;
}

void GetSeg(int L, int R, int d) {
LL cur = Seg[L][d];
for(int i=L+1; i<=R; i++) {
LL t = Seg[i][d];
Seg[i][d] += cur;
cur += t;
}
}

// u <-> v 的所有点权值+d
void add1(int u, int v, int d) {
int qd = 0, qv = d;
int ql, qr;
while(top[u] != top[v]) {
if(deep[top[u]] < deep[top[v]]) swap(u, v);
// 对应区间是p[top[u]] -> p[u];
ql = p[top[u]], qr = p[u];
SegAdd(ql, qr, qv, qd);
u = fa[top[u]];
}
if(deep[u] > deep[v]) swap(u, v);
ql = p[u], qr = p[v];
SegAdd(ql, qr, qv, qd);
}

// u <-> v 的所有边权值+d
void add2(int u, int v, int d) {
int f1 = top[u], f2 = top[v];
int qd = 1, qv = d;
int ql, qr;
while(top[u] != top[v]) {
if(deep[top[u]] < deep[top[v]]) swap(u, v);
// 对应区间是p[top[u]] -> p[u];
ql = p[top[u]], qr = p[u];
SegAdd(ql, qr, qv, qd);
u = fa[top[u]];
}
if(u == v) return ;
if(deep[u] > deep[v]) swap(u,v);
// 注意区间是p[son[u]] -> p[v]
ql = p[son[u]], qr = p[v];
SegAdd(ql, qr, qv, qd);
}

int eu[maxn], ev[maxn];
char str[10];

int main() {
int T;

scanf("%d", &T);
for(int kase=1; kase<=T; kase++) {
int n, m;
init();
printf("Case #%d:\n", kase);
scanf("%d%d", &n, &m);
for(int i=0; i<n-1; i++) {
scanf("%d%d", &eu[i], &ev[i]);
addedge(eu[i], ev[i]);
addedge(ev[i], eu[i]);
}
dfs1(1, 1, 1);
dfs2(1, 1);
memset(Seg, 0, sizeof(Seg));
for(int i=0; i<m; i++) {
int u, v, d;
scanf("%s%d%d%d", str, &u, &v, &d);
if(str[3] == '1') {
add1(u, v, d);
} else {
add2(u, v, d);
}
}
GetSeg(1, pos-1, 0);
GetSeg(1, pos-1, 1);
for(int i=1; i<=n; i++) {
LL ans = Seg[p[i]][0];
if(i > 1) putchar(' ');
P64I1(ans);
}
putchar('\n');
for(int i=0; i<n-1; i++) {
int u = eu[i], v = ev[i];
if(deep[u] < deep[v]) swap(u, v);
LL ans = Seg[p[u]][1];
if(i) putchar(' ');
P64I1(ans);
}
putchar('\n');
}

return 0;
}