http://acm.hdu.edu.cn/showproblem.php?pid=4897

 

题意：
给你一棵树,边的颜色要么为白色，要么为黑色，初始每条边为白色,有三种操作

1、将u-v链上面的所有边的颜色翻转 2、将u-v链上面所有邻接的边翻转(边上只有一个点在链上面) 3、询问u->v上面有多少黑色的边

 

树链剖分，线段树维护4个信息：

按dfs序建立线段树后，如果线段树内节点的区间为[l,r]，则此节点维护树上dfs序[l,r]内的父边的信息

父边指 点与父节点之间的边

sum0：节点的父边属于重链且颜色为白色 的边数

sum1：节点的父边属于重链且颜色为黑色 的边数

rev1：节点的父边颜色是否被操作1取反 （实际只会用到属于轻链的边）

rev2：节点的子树中，与节点直接相连的属于轻链边 是否被操作2取反

 

操作1：直接取反，交换sum0和sum1，维护标记rev1

细节：最后u和v（dep[u]<dep[v]）汇集到一条重链的时候，最后一次操作不包括u，因为点代表的是父边的信息

 

操作2：一条链的相邻边，除了对链上的点维护rev2操作外，

链最顶端的点如果是其父节点的重儿子，需要修改它的rev1

路径上每条重链最底端的点，如果它有重儿子，需要修改它重儿子的rev1

因为标记rev2只维护轻链

 

操作3：重链直接查，轻链呢？

在树链剖分往上跳的时候，跳轻链一定是只跳一条边

假设这条边连接了节点u和v，dep[u]<dep[v]

如果rev2[u]^rev2[v]^rev1[v] 为 true，则这条边为黑色

 

clj的题就是好哇！！！

 

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm>using namespace std;#define N 100001int n;int front[N],nxt[N<<1],to[N<<1],tot;int siz[N],dep[N],fa[N]; int bl[N],son[N]; int id[N],dy[N],cnt;bool big[N];int sum0[N<<2],sum1[N<<2]; bool rev1[N<<2],rev2[N<<2];int ans;void read(int &x) {x=0; char c=getchar();while(!isdigit(c)) c=getchar();while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); } }void add(int u,int v) {to[++tot]=v; nxt[tot]=front[u]; front[u]=tot;to[++tot]=u; nxt[tot]=front[v]; front[v]=tot; }void dfs1(int x) {siz[x]=1;for(int i=front[x];i;i=nxt[i])if(to[i]!=fa[x]){fa[to[i]]=x;dep[to[i]]=dep[x]+1;dfs1(to[i]);siz[x]+=siz[to[i]];} }void dfs2(int x,int top) {bl[x]=top;id[x]=++cnt;dy[cnt]=x;int y=0;for(int i=front[x];i;i=nxt[i])if(to[i]!=fa[x] && siz[to[i]]>siz[y]) y=to[i];if(y) {son[x]=y;big[y]=true;dfs2(y,top);}else return;for(int i=front[x];i;i=nxt[i])if(to[i]!=fa[x] && to[i]!=y) dfs2(to[i],to[i]); }void down1(int k) {rev1[k<<1]^=1;swap(sum0[k<<1],sum1[k<<1]);rev1[k<<1|1]^=1;swap(sum0[k<<1|1],sum1[k<<1|1]);rev1[k]^=1; }void down2(int k) {rev2[k<<1]^=1;rev2[k<<1|1]^=1;rev2[k]^=1; }void Reverse(int k,int l,int r,int opl,int opr,int ty) {if(l>=opl && r<=opr){if(ty==1){swap(sum1[k],sum0[k]);rev1[k]^=1;}else rev2[k]^=1;return;}int mid=l+r>>1;if(rev1[k]) down1(k);if(rev2[k]) down2(k);if(opl<=mid) Reverse(k<<1,l,mid,opl,opr,ty);if(opr>mid) Reverse(k<<1|1,mid+1,r,opl,opr,ty);if(ty==1){sum1[k]=sum1[k<<1]+sum1[k<<1|1];sum0[k]=sum0[k<<1]+sum0[k<<1|1];} }int get_lca(int u,int v) {while(bl[u]!=bl[v]){if(dep[bl[u]]<dep[bl[v]]) swap(u,v);u=fa[bl[u]];}return dep[u]<dep[v] ? u : v; }bool point_query(int k,int l,int r,int x,int ty) {if(l==r) return ty==1 ? rev1[k] : rev2[k];if(rev1[k]) down1(k);if(rev2[k]) down2(k);int mid=l+r>>1;if(x<=mid) return point_query(k<<1,l,mid,x,ty);return point_query(k<<1|1,mid+1,r,x,ty); }void query(int k,int l,int r,int opl,int opr) {if(l>=opl && r<=opr){ans+=sum1[k];return;}if(rev1[k]) down1(k);if(rev2[k]) down2(k);int mid=l+r>>1;if(opl<=mid) query(k<<1,l,mid,opl,opr);if(opr>mid) query(k<<1|1,mid+1,r,opl,opr); }void solve(int ty,int u,int v) {if(ty==1){while(bl[u]!=bl[v]){if(dep[bl[u]]<dep[bl[v]]) swap(u,v);Reverse(1,1,n,id[bl[u]],id[u],1);u=fa[bl[u]];}if(dep[u]>dep[v]) swap(u,v);if(u!=v) Reverse(1,1,n,id[u]+1,id[v],1);}else if(ty==2){int lca=get_lca(u,v);if(lca!=u && lca!=v) {if(big[lca]) Reverse(1,1,n,id[lca],id[lca],1);}else{if(dep[u]>dep[v]) swap(u,v);if(big[u]) Reverse(1,1,n,id[u],id[u],1);}while(bl[u]!=bl[v]){if(dep[bl[u]]<dep[bl[v]]) swap(u,v);if(son[u]) Reverse(1,1,n,id[son[u]],id[son[u]],1);Reverse(1,1,n,id[bl[u]],id[u],2);u=fa[bl[u]];}if(dep[u]>dep[v]) swap(u,v); if(son[v]) Reverse(1,1,n,id[son[v]],id[son[v]],1);Reverse(1,1,n,id[u],id[v],2);}else{ ans=0; while(bl[u]!=bl[v]){if(dep[bl[u]]<dep[bl[v]]) swap(u,v);query(1,1,n,id[bl[u]],id[u]);ans+=point_query(1,1,n,id[bl[u]],2)^point_query(1,1,n,id[fa[bl[u]]],2)^point_query(1,1,n,id[bl[u]],1);u=fa[bl[u]];}if(dep[u]>dep[v]) swap(u,v);if(u!=v) query(1,1,n,id[u]+1,id[v]);printf("%d\n",ans);} } void build(int k,int l,int r) {sum0[k]=sum1[k]=0;rev1[k]=rev2[k]=false;if(l==r) {sum0[k]=big[dy[l]];return;}int mid=l+r>>1;build(k<<1,l,mid);build(k<<1|1,mid+1,r);sum0[k]=sum0[k<<1]+sum0[k<<1|1]; }void clear() {tot=cnt=0;memset(front,0,sizeof(front));memset(son,0,sizeof(son));memset(big,false,sizeof(big)); }int main() {freopen("data.in","r",stdin);freopen("my.out","w",stdout);int T;read(T);int u,v;int ty,m,lca;while(T--){clear();read(n);for(int i=1;i<n;++i){read(u); read(v);add(u,v);}dfs1(1);dfs2(1,1);build(1,1,n);read(m);while(m--){read(ty); read(u); read(v);solve(ty,u,v);}}return 0; }

 

Little Devil I

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1087    Accepted Submission(s): 378

Problem Description There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

The devil likes to make thing in chaos. This kingdom’s road system is like simply a tree(connected graph without cycle). A road has a color of black or white. The devil often wants to make some change of this system.

In details, we call a path on the tree from a to b consists of vertices lie on the shortest simple path between a and b. And we say an edge is on the path if both its two endpoints is in the path, and an edge is adjacent to the path if exactly one endpoint of it is in the path.

Sometimes the devil will ask you to reverse every edge’s color on a path or adjacent to a path.

The king’s daughter, WJMZBMR, is also a cute loli, she is surprised by her father’s lolicon-like behavior. As she is concerned about the road-system’s status, sometimes she will ask you to tell there is how many black edge on a path.

Initially, every edges is white.

 

Input The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains an integer n, which is the size of the tree. The vertices be indexed from 1.
On the next n-1 lines, each line contains two integers a,b, denoting there is an edge between a and b.
The next line contains an integer Q, denoting the number of the operations.
On the next Q lines, each line contains three integers t,a,b. t=1 means we reverse every edge’s color on path a to b. t=2 means we reverse every edge’s color adjacent to path a to b. t=3 means we query about the number of black edge on path a to b.

T<=5.
n,Q<=10^5.
Please use scanf,printf instead of cin,cout,because of huge input.

 

Output For each t=3 operation, output the answer in one line.

 

Sample Input 1 10 2 1 3 1 4 1 5 1 6 5 7 4 8 3 9 5 10 6 10 2 1 6 1 3 8 3 8 10 2 3 4 2 10 8 2 4 10 1 7 6 2 7 3 2 1 4 2 10 10

 

Sample Output 3 Hint reverse color means change from white to black or vice virsa.

 

Author WJMZBMR

转载于:https://www.cnblogs.com/TheRoadToTheGold/p/8456845.html

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