P3038 [USACO11DEC]牧草种植Grass Planting
题目描述
Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on the roads between pastures. Farmer John loves Bessie very much, and today he is finally going to plant grass on the roads. He will do so using a procedure consisting of M steps (1 <= M <= 100,000).
At each step one of two things will happen:
- FJ will choose two pastures, and plant a patch of grass along each road in between the two pastures, or,
- Bessie will ask about how many patches of grass on a particular road, and Farmer John must answer her question.
Farmer John is a very poor counter -- help him answer Bessie‘s questions!
给出一棵n个节点的树，有m个操作，操作为将一条路径上的边权加一或询问某条边的权值。
输入输出格式
输入格式：
- Line 1: Two space-separated integers N and M
- Lines 2..N: Two space-separated integers describing the endpoints of a road.
- Lines N+1..N+M: Line i+1 describes step i. The first character of the line is either P or Q, which describes whether or not FJ is planting grass or simply querying. This is followed by two space-separated integers A_i and B_i (1 <= A_i, B_i <= N) which describe FJ‘s action or query.
输出格式：
- Lines 1..???: Each line has the answer to a query, appearing in the same order as the queries appear in the input.
输入输出样例
输入样例#1：
4 6 1 4 2 4 3 4 P 2 3 P 1 3 Q 3 4 P 1 4 Q 2 4 Q 1 4
输出样例#1：
2 1 2
树剖。。
1 #include <ctype.h>
2 #include <cstdio>
3 #include <queue>
4
5 const int MAXN=100010;
6
7 int n,m,inr;
8
9 int dfn[MAXN],dep[MAXN],id[MAXN],fa[MAXN];
10 int top[MAXN],son[MAXN],siz[MAXN],rank[MAXN];
11
12 struct SegmentTree {
13 int l,r;
14 int tag;
15 int sum;
16 };
17 SegmentTree t[MAXN<<2];
18
19 struct node {
20 int to;
21 int next;
22 };
23 node e[MAXN<<1];
24
25 int head[MAXN],tot;
26
27 inline void read(int&x) {
28 int f=1;register char c=getchar();
29 for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
30 for(;isdigit(c);x=x*10+c-48,c=getchar());
31 x=x*f;
32 }
33
34 inline void add(int x,int y) {
35 e[++tot].to=y;
36 e[tot].next=head[x];
37 head[x]=tot;
38 }
39
40 void Dfs_1(int now,int f) {
41 dep[now]=dep[f]+1;
42 siz[now]=1;
43 fa[now]=f;
44 for(int i=head[now];i;i=e[i].next) {
45 int to=e[i].to;
46 if(to==f) continue;
47 Dfs_1(to,now);
48 siz[now]+=siz[to];
49 if(son[now]==-1||siz[son[now]]<siz[to]) son[now]=to;
50 }
51 return;
52 }
53
54 void Dfs_2(int now,int tp) {
55 top[now]=tp;
56 id[now]=++inr;
57 rank[inr]=now;
58 if(son[now]==-1) return;
59 Dfs_2(son[now],tp);
60 for(int i=head[now];i;i=e[i].next) {
61 int to=e[i].to;
62 if(to==son[now]||to==fa[now]) continue;
63 Dfs_2(to,to);
64 }
65 return;
66 }
67
68 inline void swap(int&x,int&y) {
69 int t=x;x=y;y=t;
70 return;
71 }
72
73 inline void down(int now) {
74 t[now<<1].tag+=t[now].tag;
75 t[now<<1].sum+=(t[now<<1].r-t[now<<1].l+1)*t[now].tag;
76 t[now<<1|1].tag+=t[now].tag;
77 t[now<<1|1].sum+=(t[now<<1|1].r-t[now<<1|1].l+1)*t[now].tag;
78 t[now].tag=0;
79 }
80
81 void build_tree(int now,int l,int r) {
82 t[now].l=l;t[now].r=r;
83 if(l==r) return;
84 int mid=(l+r)>>1;
85 build_tree(now<<1,l,mid);
86 build_tree(now<<1|1,mid+1,r);
87 }
88
89 void modify(int now,int l,int r) {
90 if(l<=t[now].l&&r>=t[now].r) {
91 ++t[now].tag;
92 t[now].sum+=t[now].r-t[now].l+1;
93 return;
94 }
95 if(t[now].tag) down(now);
96 int mid=(t[now].l+t[now].r)>>1;
97 if(l<=mid) modify(now<<1,l,r);
98 if(r>mid) modify(now<<1|1,l,r);
99 t[now].sum=t[now<<1].sum+t[now<<1|1].sum;
100 }
101
102 int query(int now,int l,int r) {
103 int ans=0;
104 if(l<=t[now].l&&r>=t[now].r) return t[now].sum;
105 if(t[now].tag) down(now);
106 int mid=(t[now].l+t[now].r)>>1;
107 if(l<=mid) ans+=query(now<<1,l,r);
108 if(r>mid) ans+=query(now<<1|1,l,r);
109 return ans;
110 }
111
112 inline void Pre_query(char c,int x,int y) {
113 int ans=0;
114 while(top[x]!=top[y]) {
115 if(dep[top[x]]<dep[top[y]]) swap(x,y);
116 if(c==‘P‘) modify(1,id[top[x]],id[x]);
117 else ans+=query(1,id[top[x]],id[x]);
118 x=fa[top[x]];
119 }
120 if(dep[x]>dep[y]) swap(x,y);
121 if(c==‘P‘) modify(1,id[x]+1,id[y]);
122 else ans+=query(1,id[x]+1,id[y]),printf("%d\n",ans);
123 return;
124 }
125
126 int hh() {
127 char s[5];
128 read(n);read(m);
129 for(int i=1;i<=n;++i) son[i]=-1;
130 int t=n-1;
131 for(int x,y;t--;) {
132 read(x);read(y);
133 add(x,y);add(y,x);
134 }
135 Dfs_1(1,0);
136 Dfs_2(1,1);
137 build_tree(1,1,inr);
138 for(int x,y;m--;) {
139 scanf("%s",s);read(x);read(y);
140 Pre_query(s[0],x,y);
141 }
142 return 0;
143 }
144
145 int sb=hh();
146 int main() {;}
代码
时间： 2024-10-09 16:02:10