2019icpc Xi'an tournament J And And And (tree dp)

Topic link: https: //nanti.jisuanke.com/t/39277

Meaning of the questions: have given an edge weight of the tree, find all paths that contain simple XOR and simple path for the total number of 0's and.

Ideas:

　　First, the exclusive OR and val [I] XOR to zero on this limitation, we get all points to the root node through a path dfs, if the value val is equal to two nodes, with the path between them meet different or is zero. SZ [i] i in a subtree rooted size.

　　Secondly, to meet the XOR of 0 and two points u, v, two cases to consider:

　　　　1. u, v on a different chain, the contribution of this path is a sz [u] * sz [v], by the size of each record and map weights can be obtained. First of all we point this process again.

　　　　2. u, v on one strand, this contribution is a path (n-sz [u1]) * sz [v], u1 is the first child of v u u on this strand. Because the previous step we added all cases, it must first subtract sz [u] * sz [v], to record on the same or different chain of size and value val by mp1, because the record is the same chain, when the process all child nodes of a node to back the update, i.e. preceded by subtracting the size. After adding the Save (n-sz [u1]) * sz [v], the information is recorded by mp2, updated every processed back a child node, because the longer strand.

AC Code:

#include<cstdio>
#include<algorithm>
#include<unordered_map>
using namespace std;

const int maxn=1e5+5;
const int MOD=1e9+7;
typedef long long LL;
typedef unordered_map<LL,LL> ump;
struct node{
    int v,nex;
    LL w;
}edge[maxn];

int n,head[maxn],cnt,sz[maxn];
LL val[maxn],ans;
ump sum,mp1,mp2;

void adde(int u,int v,LL w){
    edge[++cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].nex=head[u];
    head[u]=cnt;
}

void dfs1(int u,LL va){
    val[u]=va,sz[u]=1;
    for(int i=head[u];i;i=edge[i].nex){
        int v=edge[i].v;
        LL w=edge[i].w;
        dfs1(v,va^w);
        sz[u]+=sz[v];
    }
}

void dfs2(int u){
    ans=(ans+sum[val[u]]*sz[u])%MOD;
    sum[val[u]]=(sum[val[u]]+sz[u])%MOD;
    for(int i=head[u];i;i=edge[i].nex)
        dfs2(edge[i].v);
}

void dfs3(int u){
    ans=(ans-mp1[val[u]]*sz[u]+mp2[val[u]]*sz[u]+MOD)%MOD;
    mp1[val[u]]=(mp1[val[u]]+sz[u])%MOD;
    for(int i=head[u];i;i=edge[i].nex){
        int v=edge[i].v;
        LL w=edge[i].w;
        mp2[val[u]]=(mp2[val[u]]+n-sz[v]+MOD)%MOD;
        dfs3(v);
        mp2[val[u]]=(mp2[val[u]]-n+sz[v]+MOD)%MOD;
    }
    mp1[val[u]]=(mp1[val[u]]-sz[u]+MOD)%MOD;
}

int main(){
    scanf("%d",&n);
    for(int i=2;i<=n;++i){
        int u;LL w;
        scanf("%d%lld",&u,&w);
        adde(u,i,w);
    }
    dfs1(1,0);
    dfs2(1);
    dfs3(1);
    printf("%lld\n",ans);
    return 0;
}