Tree
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 567 Accepted Submission(s): 121
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.
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)
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.
題意:
一棵樹,初始點權和邊權都爲0。
兩種操作
add1 u v d : u -> v之間的所有點權值+d
add2 u v d : u -> v之間的所有邊權值+d
樹鏈剖分入門模板題
比賽的時候才學的竟然很順利的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所在的重鏈的頂端節點
int fa[maxn]; // 父親節點
int deep[maxn]; // 深度
int num[maxn]; // num[v]表示以v爲根的子樹的節點數
int p[maxn]; // p[v]表示v與其父親節點的連邊在線段樹中的位置
int fp[maxn]; // 和p數組相反
int son[maxn]; // 重兒子
int pos; // 對應區間計數器
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當前節點, pre爲父節點, d爲深度
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爲當前點所在重鏈的頂節點
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);
}
}
// 區間+d,最後求每個點的值
// 離線標記法
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 的所有點權值+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);
// 對應區間是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 的所有邊權值+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);
// 對應區間是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);
// 注意區間是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;
}