HDU Aragorn's Story -树链剖分

Posted 2020-10-03 心之所向 素履以往

                                                   HDU Aragorn\'s Story 

 

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.

 

 

Source

2011 Multi-University Training Contest 13 - Host by HIT

 

不说话，先上图

 

 

惨痛的教训。。。。

以后再也不输入一个字符了。。

 

  1 #include <cstdio>
  2 #include <ctype.h>
  3 #include <cstring>
  4 
  5 const int MAXN=50002;
  6 
  7 int n,m,q,inr;
  8 
  9 int son[MAXN],siz[MAXN],fa[MAXN],id[MAXN];
 10 int rank[MAXN],a[MAXN],dep[MAXN],top[MAXN];
 11 
 12 struct SegmentTree {
 13     int l,r;
 14     int tag,sum;
 15 };
 16 SegmentTree tr[MAXN<<2];
 17 
 18 struct Edge {
 19     int to;
 20     int next;
 21 };
 22 Edge e[MAXN<<1];
 23 
 24 int head[MAXN<<1],tot;
 25 
 26 inline void read(int&x) {
 27     int f=1;register char c=getchar();
 28     for(x=0;!isdigit(c);c==\'-\'&&(f=-1),c=getchar());
 29     for(;isdigit(c);x=x*10+c-48,c=getchar());
 30     x=x*f;
 31 }
 32 
 33 inline void add(int x,int y) {
 34     e[++tot].to=y;
 35     e[tot].next=head[x];
 36     head[x]=tot;
 37 }
 38 
 39 inline void init() {
 40     for(int i=0;i<(inr<<2);i++) tr[i].tag=0;
 41     memset(head,-1,sizeof head);
 42     memset(son,-1,sizeof son);
 43     memset(dep,0,sizeof dep);
 44     memset(fa,0,sizeof fa);
 45     tot=0;inr=0;
 46 }
 47 
 48 inline void swap(int&a,int&b) {
 49     int t=a;a=b;b=t;
 50 }
 51 
 52 void Dfs_1(int now,int f) {
 53     dep[now]=dep[f]+1;
 54     fa[now]=f;
 55     siz[now]=1;
 56     for(int i=head[now];i!=-1;i=e[i].next) {
 57         int to=e[i].to;
 58         if(to==f) continue;
 59         Dfs_1(to,now);
 60         siz[now]+=siz[to];
 61         if(son[now]==-1||siz[son[now]]<son[to]) son[now]=to;
 62     }
 63     return;
 64 }
 65 
 66 void Dfs_2(int now,int tp) {
 67     id[now]=++inr;
 68     rank[inr]=now;
 69     top[now]=tp;
 70     if(son[now]==-1) return;
 71     Dfs_2(son[now],tp);
 72     for(int i=head[now];i!=-1;i=e[i].next) {
 73         int to=e[i].to;
 74         if(to==fa[now]||to==son[now]) continue;
 75         Dfs_2(to,to);
 76     }
 77     return;
 78 }
 79 
 80 inline void down(int now) {
 81     tr[now<<1].tag+=tr[now].tag;
 82     tr[now<<1|1].tag+=tr[now].tag;
 83     tr[now<<1].sum+=(tr[now<<1].r-tr[now<<1].l+1)*tr[now].tag;
 84     tr[now<<1|1].sum+=(tr[now<<1|1].r-tr[now<<1|1].l+1)*tr[now].tag;
 85     tr[now].tag=0;
 86 }
 87 
 88 void build_tree(int now,int l,int r) {
 89     tr[now].l=l;tr[now].r=r;
 90     if(l==r) {
 91         tr[now].sum=a[rank[l]];
 92         return;
 93     }
 94     int mid=(l+r)>>1;
 95     build_tree(now<<1,l,mid);
 96     build_tree(now<<1|1,mid+1,r);
 97     tr[now].sum=tr[now<<1].sum+tr[now<<1|1].sum;
 98     return;
 99 }
100 
101 void modify(int now,int l,int r,int v) {
102     if(l<=tr[now].l&&r>=tr[now].r) {
103         tr[now].tag+=v;
104         tr[now].sum+=(tr[now].r-tr[now].l+1)*v;
105         return;
106     }
107     if(tr[now].tag) down(now);
108     int mid=(tr[now].l+tr[now].r)>>1;
109     if(l<=mid) modify(now<<1,l,r,v);
110     if(r>mid) modify(now<<1|1,l,r,v);
111     tr[now].sum=tr[now<<1].sum+tr[now<<1|1].sum;
112     return;
113 }
114 
115 int query(int now,int pos) {
116     if(tr[now].l==tr[now].r) return tr[now].sum;
117     if(tr[now].tag) down(now);
118     int mid=(tr[now].l+tr[now].r)>>1;
119     if(pos<=mid) return query(now<<1,pos);
120     else return query(now<<1|1,pos);
121 }
122 
123 inline void Pre(int x,int y,int v) {
124     while(top[x]!=top[y]) {
125         if(dep[top[x]]<dep[top[y]]) swap(x,y);
126         modify(1,id[top[x]],id[x],v);
127         x=fa[top[x]];
128     }
129     if(dep[x]>dep[y]) swap(x,y);
130     modify(1,id[x],id[y],v);
131 }
132 
133 int hh() {
134     while(scanf("%d%d%d",&n,&m,&q)!=EOF) {
135         init();
136         char c[10];int x,y,v;
137         for(int i=1;i<=n;++i) read(a[i]);
138         for(int i=1;i<=m;++i) 
139           read(x),read(y),add(x,y),add(y,x);
140         Dfs_1(1,0);Dfs_2(1,1);
141         build_tree(1,1,inr);
142         for(int i=1;i<=q;++i) {
143             scanf("%s",c);read(x);
144             if(c[0]==\'I\') {
145                 read(y);read(v);
146                 Pre(x,y,v);
147             }
148             else if(c[0]==\'D\') {
149                 read(y);read(v);
150                 Pre(x,y,-v);
151             }
152             else {
153                 int ans=query(1,id[x]);
154                 printf("%d\\n",ans);
155             }
156         }
157     }
158     return 0;
159 }
160 
161 int sb=hh();
162 int main() {;}

代码