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.
题意：有n个camp，初始时每个camp的人数给定，有些camp之间有边连接，并且恰好是一棵树。
现在支持3种操作：
1.I a b c :把a到b的路径(包括a，b)经过的每一个camp的人数+c
2.D a b c :把a到b的路径(包括a，b)经过的每一个camp的人数-c
3.Q a:查询a这个camp有多少个人。
明显，树链剖分+线段树维护
这是我做的第一道树链剖分的题目。
其实树链剖分就是把树上的节点重新编号为1~n,然后操作就可以用数据结构维护了。
has[u]表示树上节点u重新编号后为has[u],类似于哈希hash。
遇上一个问题：
要操作树上的(a,b)这个路径的所有节点，但是这个路径不等于重新编号后的[has(a),has(b)]
而是由几个区间的和。
所以我们就是要找到，这条路径在has后是由哪部分区间的和组成的。
只要找到这2个点的在树上的最近公共祖先就可以了。
怎么找呢？
看代码里面的change() 函数。
不断边更新，边走完这条链。
1 #pragma comment (linker,"/STACK:1024000000,1024000000")
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5
6 using namespace std;
7
8 const int maxn=50000+5;
9
10 #define ll long long
11
12 struct Edge
13 {
14 int to,next;
15 }edge[maxn<<1];
16 int head[maxn];
17 int tot,num;
18 int n;
19
20 void addedge(int u,int v)
21 {
22 edge[tot].to=v;
23 edge[tot].next=head[u];
24 head[u]=tot++;
25 }
26
27 int fa[maxn];
28 int siz[maxn];
29 int top[maxn];
30 int dep[maxn];
31 int son[maxn];
32 int val[maxn];
33 int has[maxn];
34 int ran[maxn];
35
36 void dfs1(int u,int father,int d)
37 {
38 fa[u]=father;
39 siz[u]=1;
40 dep[u]=d;
41 for(int i=head[u];~i;i=edge[i].next)
42 {
43 int v=edge[i].to;
44 if(v==father)
45 continue;
46 dfs1(v,u,d+1);
47 siz[u]+=siz[v];
48 if(son[u]==-1||siz[v]>siz[son[u]])
49 {
50 son[u]=v;
51 }
52 }
53 }
54
55 void dfs2(int u,int tp)
56 {
57 top[u]=tp;
58 has[u]=num;
59 ran[num]=u;
60 num++;
61 if(son[u]==-1)
62 return ;
63 dfs2(son[u],tp);
64
65 for(int i=head[u];~i;i=edge[i].next)
66 {
67 int v=edge[i].to;
68 if(v==fa[u]||v==son[u])
69 continue;
70 dfs2(v,v);
71 }
72 }
73
74 #define lson l,m,rt<<1
75 #define rson m+1,r,rt<<1|1
76
77 ll sum[maxn<<2];
78 ll col[maxn<<2];
79
80 void pushup(int rt)
81 {
82 sum[rt]=sum[rt<<1]+sum[rt<<1|1];
83 }
84
85 void pushdown(int l,int r,int rt)
86 {
87 if(col[rt])
88 {
89 int m=(l+r)>>1;
90 col[rt<<1]+=col[rt];
91 col[rt<<1|1]+=col[rt];
92 sum[rt<<1]+=col[rt]*(m-l+1);
93 sum[rt<<1|1]+=col[rt]*(r-m);
94 col[rt]=0;
95 }
96 }
97
98 void build(int l,int r,int rt)
99 {
100 if(l==r)
101 {
102 sum[rt]=(ll)val[ran[l]];
103 return ;
104 }
105
106 int m=(l+r)>>1;
107 build(lson);
108 build(rson);
109 pushup(rt);
110 }
111
112 void update(int L,int R,int add,int l,int r,int rt)
113 {
114 if(L<=l&&R>=r)
115 {
116 col[rt]+=(ll)add;
117 sum[rt]+=(ll)add*(r-l+1);
118 return ;
119 }
120
121 int m=(l+r)>>1;
122
123 pushdown(l,r,rt);
124
125 if(L<=m)
126 update(L,R,add,lson);
127 if(R>m)
128 update(L,R,add,rson);
129 pushup(rt);
130 }
131
132 ll query(int p,int l,int r,int rt)
133 {
134 if(l==r)
135 {
136 return sum[rt];
137 }
138
139 int m=(l+r)>>1;
140 pushdown(l,r,rt);
141
142 ll ret=0;
143
144 if(p<=m)
145 ret=query(p,lson);
146 else
147 ret=query(p,rson);
148
149 pushup(rt);
150
151 return ret;
152 }
153
154 void init()
155 {
156 memset(head,-1,sizeof(head));
157 memset(son,-1,sizeof(son));
158 memset(col,0,sizeof(col));
159 tot=1;
160 num=1;
161 }
162
163 void change(int u,int v,int w)
164 {
165 while(top[u]!=top[v])
166 {
167 if(dep[top[u]]<dep[top[v]])
168 {
169 swap(u,v);
170 }
171 update(has[top[u]],has[u],w,1,n,1);
172 u=fa[top[u]];
173 }
174 if(dep[u]>dep[v])
175 swap(u,v);
176 update(has[u],has[v],w,1,n,1);
177 }
178
179 int main()
180 {
181 while(scanf("%d",&n)!=EOF)
182 {
183 init();
184
185 int m,q;
186 scanf("%d%d",&m,&q);
187
188 for(int i=1;i<=n;i++)
189 scanf("%d",&val[i]);
190
191 for(int i=1;i<=m;i++)
192 {
193 int u,v;
194 scanf("%d%d",&u,&v);
195 addedge(u,v);
196 addedge(v,u);
197 }
198
199 dfs1(1,1,1);
200 dfs2(1,1);
201
202 build(1,n,1);
203
204 for(int i=1;i<=q;i++)
205 {
206 char str[5];
207 scanf("%s",&str);
208 if(str[0]==‘I‘||str[0]==‘D‘)
209 {
210 int uu,vv,ww;
211 scanf("%d%d%d",&uu,&vv,&ww);
212
213 if(str[0]==‘D‘)
214 ww=-ww;
215
216 change(uu,vv,ww);
217
218 }
219 else
220 {
221 int uu;
222 scanf("%d",&uu);
223 printf("%lld\n",query(has[uu],1,n,1));
224 }
225 }
226
227 }
228
229 return 0;
230 }
HDU 2966 Aragorn's Story 树链剖分第一题 基础题