Aragorn’s Story
Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15842 Accepted Submission(s): 4159
Problem Description
Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.
Input
Multiple test cases, process to the end of input.
For each case, The first line contains three integers N, M, P which means there will be N(1 ≤ N ≤ 50000) camps, M(M = N-1) edges and P(1 ≤ P ≤ 100000) operations. The number of camps starts from 1.
The next line contains N integers A1, A2, …AN(0 ≤ Ai ≤ 1000), means at first in camp-i has Ai enemies.
The next M lines contains two integers u and v for each, denotes that there is an edge connects camp-u and camp-v.
The next P lines will start with a capital letter ‘I’, ‘D’ or ‘Q’ for each line.
‘I’, followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, increase K soldiers to these camps.
‘D’, followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, decrease K soldiers to these camps.
‘Q’, followed by one integer C, which is a query and means Aragorn wants to know the number of enemies in camp C at that time.
Output
For each query, you need to output the actually number of enemies in the specified camp.
Sample Input
3 2 5
1 2 3
2 1
2 3
I 1 3 5
Q 2
D 1 2 2
Q 1
Q 3
Sample Output
7
4
8
Hint
1.The number of enemies may be negative.
2.Huge input, be careful.
树链剖分点权的模板题：
#include <bits/stdc++.h>
using namespace std;
/*
* 基于点权,查询单点值,修改路径的上的点权
*/
const int MAXN = 50010;
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对应的位置
int fp[MAXN]; // fp和p数组相反
int son[MAXN]; // 重儿子
int pos;
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
pos = 1; // 使用树状数组,编号从头1开始
memset(son, -1, sizeof(son));
return ;
}
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
return ;
}
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;
}
}
}
return ;
}
void getpos(int u, int sp)
{
top[u] = sp;
p[u] = pos++;
fp[p[u]] = u;
if (son[u] == -1)
{
return ;
}
getpos(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])
{
getpos(v, v);
}
}
return ;
}
//树状数组
int lowbit(int x)
{
return x & (-x);
}
int c[MAXN];
int n;
int sum(int i)
{
int s = 0;
while (i > 0)
{
s += c[i];
i -= lowbit(i);
}
return s;
}
void add(int i, int val)
{
while (i <= n)
{
c[i] += val;
i += lowbit(i);
}
return ;
}
void Change(int u, int v, int val) // u->v的路径上点的值改变val
{
int f1 = top[u], f2 = top[v];
while (f1 != f2)
{
if (deep[f1] < deep[f2])
{
swap(f1, f2);
swap(u, v);
}
add(p[f1], val);
add(p[u] + 1, -val);
u = fa[f1];
f1 = top[u];
}
if (deep[u] > deep[v])
{
swap(u, v);
}
add(p[u], val);
add(p[v] + 1, -val);
return ;
}
int a[MAXN];
int main()
{
int M, P;
while (scanf("%d%d%d", &n, &M, &P) == 3)
{
int u, v;
int C1, C2, K;
char op[10];
init();
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
}
while(M--)
{
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
}
dfs1(1, 0, 0);
getpos(1, 1);
memset(c, 0, sizeof(c));
for (int i = 1; i <= n; i++)
{
add(p[i],a[i]);
add(p[i] + 1, -a[i]);
}
while (P--)
{
scanf("%s", op);
if (op[0] == 'Q')
{
scanf("%d", &u);
printf("%d
", sum(p[u]));
}
else
{
scanf("%d%d%d", &C1, &C2, &K);
if (op[0] == 'D')
{
K = -K;
}
Change(C1, C2, K);
}
}
}
return 0;
}