Tree
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 13156 | Accepted: 3358 |
题目链接
http://poj.org/problem?id=3237
Description
You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:
CHANGE i v | Change the weight of the ith edge to v |
|---|---|
NEGATE a b |
Negate the weight of every edge on the path from a to b |
QUERY a b |
Find the maximum weight of edges on the path from a to b |
Input
The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.
Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.
Output
For each “QUERY” instruction, output the result on a separate line.
Sample Input
1
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
Sample Output
1
3
题解
树练剖分的模板题，单点修改，区间查询，还有区间取反（就是变成相反数。。。）。
注意一下单点的lazy标记处理就好了，我是每个区间权值由子区间和lazy标记组成，
然后叶子节点lazy区间用一次就消失。
代码
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 using namespace std; 6 #define ll long long 7 #define N 100050 8 #define INF 0x7f7f7f7f 9 int n,bh[N],w[N]; 10 struct Tree{int l,r,lazy,max,min;}tr[N<<2]; 11 struct Edge{int from,to,val,id,s;}edges[N<<1]; 12 int tot,last[N]; 13 int cnt,fa[N],size[N],dp[N],son[N],rk[N],kth[N],top[N]; 14 15 template<typename T>void read(T&x) 16 { 17 ll k=0; char c=getchar(); 18 x=0; 19 while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar(); 20 if (c==EOF)exit(0); 21 while(isdigit(c))x=x*10+c-'0',c=getchar(); 22 x=k?-x:x; 23 } 24 void AddEdge(int x,int y,int z,int id) 25 { 26 edges[++tot]=Edge{x,y,z,id,last[x]}; 27 last[x]=tot; 28 } 29 void read_char(char &c) 30 {while(!isalpha(c=getchar())&&c!=EOF);} 31 void push_up(int x) 32 { 33 int len=(tr[x].r-tr[x].l+1); 34 if (len>1) 35 { 36 tr[x].max=max(tr[x<<1].max,tr[x<<1|1].max); 37 tr[x].min=min(tr[x<<1].min,tr[x<<1|1].min); 38 } 39 if (tr[x].lazy==-1) 40 { 41 tr[x].max*=-1; 42 tr[x].min*=-1; 43 swap(tr[x].max,tr[x].min); 44 } 45 if (len==1)tr[x].lazy=1; 46 } 47 void push_down(int x) 48 { 49 tr[x<<1].lazy*=tr[x].lazy; 50 tr[x<<1|1].lazy*=tr[x].lazy; 51 push_up(x<<1); 52 push_up(x<<1|1); 53 tr[x].lazy=1; 54 } 55 void bt(int x,int l,int r) 56 { 57 tr[x].l=l; tr[x].r=r; tr[x].lazy=1; 58 if (l==r) 59 { 60 tr[x].max=tr[x].min=w[kth[l]]; 61 return; 62 } 63 int mid=(l+r)>>1; 64 bt(x<<1,l,mid); 65 bt(x<<1|1,mid+1,r); 66 push_up(x); 67 } 68 void update(int x,int p,int tt) 69 { 70 if (p<=tr[x].l&&tr[x].r<=p) 71 { 72 tr[x].max=tr[x].min=tt; 73 tr[x].lazy=1; 74 return; 75 } 76 int mid=(tr[x].l+tr[x].r)>>1; 77 push_down(x); 78 if (p<=mid)update(x<<1,p,tt); 79 if (mid<p)update(x<<1|1,p,tt); 80 push_up(x); 81 } 82 int neg(int x,int l,int r,int tt) 83 { 84 if (l<=tr[x].l&&tr[x].r<=r) 85 { 86 tr[x].lazy*=tt; 87 push_up(x); 88 return tr[x].max; 89 } 90 int mid=(tr[x].l+tr[x].r)>>1,ans=-INF; 91 push_down(x); 92 if (l<=mid)ans=max(ans,neg(x<<1,l,r,tt)); 93 if (mid<r)ans=max(ans,neg(x<<1|1,l,r,tt)); 94 push_up(x); 95 return ans; 96 } 97 void dfs1(int x,int pre) 98 { 99 fa[x]=pre; 100 dp[x]=dp[pre]+1; 101 size[x]=1; 102 son[x]=0; 103 for(int i=last[x];i;i=edges[i].s) 104 { 105 Edge &e=edges[i]; 106 if (e.to==pre)continue; 107 w[e.to]=e.val; 108 bh[e.id]=e.to; 109 dfs1(e.to,x); 110 size[x]+=size[e.to]; 111 if (size[e.to]>size[son[x]])son[x]=e.to; 112 } 113 } 114 void dfs2(int x,int y) 115 { 116 rk[x]=++cnt; 117 kth[cnt]=x; 118 top[x]=y; 119 if (son[x]==0)return; 120 dfs2(son[x],y); 121 for(int i=last[x];i;i=edges[i].s) 122 { 123 Edge &e=edges[i]; 124 if (e.to==fa[x]||e.to==son[x])continue; 125 dfs2(e.to,e.to); 126 } 127 } 128 int get_max(int x,int y,int tt) 129 { 130 int fx=top[x],fy=top[y],ans=-INF; 131 while(fx!=fy) 132 { 133 if (dp[fx]<dp[fy])swap(x,y),swap(fx,fy); 134 ans=max(ans,neg(1,rk[fx],rk[x],tt)); 135 x=fa[fx];fx=top[x]; 136 } 137 if (dp[x]<dp[y])swap(x,y); 138 ans=max(ans,neg(1,rk[y]+1,rk[x],tt)); 139 return ans; 140 } 141 void work() 142 { 143 read(n); 144 for(int i=1;i<=n-1;i++) 145 { 146 int x,y,z; 147 read(x); read(y); read(z); 148 AddEdge(x,y,z,i); 149 AddEdge(y,x,z,i); 150 } 151 dfs1(1,0); 152 dfs2(1,1); 153 bt(1,1,n); 154 while(1) 155 { 156 char id; int x,y; 157 read_char(id); 158 if (id=='D') break; 159 read(x); read(y); 160 if (id=='C') update(1,rk[bh[x]],y); 161 if (id=='N') get_max(x,y,-1); 162 if (id=='Q') printf("%d ",get_max(x,y,1)); 163 } 164 } 165 void clear() 166 { 167 cnt=0; tot=0; 168 memset(last,0,sizeof(last)); 169 } 170 int main() 171 { 172 #ifndef ONLINE_JUDGE 173 freopen("aa.in","r",stdin); 174 #endif 175 int q; 176 read(q); 177 while(q--) 178 { 179 clear(); 180 work(); 181 } 182 }