GFOJ510树上路径

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Problem 510: 树上路径
Time Limit: 3000 ms
Memory Limit: 512000 KB

Problem Description

这里写图片描述

练一练点分……
dep表示点到重心距离，mx表示点到重心最大值，对于mx排序后乱搞……
表示拍了很久发现犯了个**错误……

#include <cstdio>#include <algorithm>#define maxn 100000+10using namespace std;int num,x,y,dep[maxn],_mx[maxn],rot,ans,now_sz,vis[maxn],sz[maxn],f[maxn],g[10000000+10],n,p,head[maxn],m,a[maxn];struct xx{    int v,next;}b[maxn<<1];struct yy{    int dep,mx;}c[maxn];void add(int u,int v){    b[++m]=(xx){v,head[u]};    head[u]=m;}void getrot(int x,int fa){    f[x]=0;sz[x]=1;    for (int i=head[x];i;i=b[i].next)    {        int v=b[i].v;        if (v==fa||vis[v]) continue;        getrot(v,x);        sz[x]+=sz[v];        f[x]=max(f[x],sz[v]);    }    f[x]=max(now_sz-sz[x],f[x]);    rot=f[x]<f[rot]?x:rot;}void dfs(int x,int fa){    for (int i=head[x];i;i=b[i].next)    {        int v=b[i].v;        if (vis[v]||v==fa)continue;        /*dep[v]=(dep[x]+a[x])%p;        _mx[v]=max(_mx[v],a[x]);*/        dep[v]=(dep[x]+a[v])%p;        _mx[v]=max(_mx[x],a[v]);        c[++num]=(yy){dep[v],_mx[v]};        dfs(v,x);    }}bool cmp(const yy&a,const yy&b){    return a.mx<b.mx;}int cal(int x,int f){    int tmp=0;    num=0;    dep[x]=f==-1?a[x]%p:(a[x]+f)%p;    _mx[x]=f==-1?a[x]:max(a[x],f);    c[++num]=(yy){dep[x],_mx[x]};    dfs(x,0);    sort(c+1,c+num+1,cmp);    for (int i=1;i<=num;++i)    {        int hh=(p+c[i].mx+(f==-1?a[x]:f)-c[i].dep)%p;//(f==-1?a[x]:f)不要写成a[x]!!!         tmp+=g[hh];        ++g[c[i].dep];    }    for (int i=1;i<=num;++i) --g[c[i].dep];    return tmp;}void work(int x){    vis[x]=1;    ans+=cal(x,-1);    for (int i=head[x];i;i=b[i].next)    {        int v=b[i].v;        if (vis[v])continue;        ans-=cal(v,a[x]);        now_sz=sz[v],rot=0;        getrot(v,0);        work(rot);    }}int main(){    freopen("data.out","r",stdin);    freopen("wa.out","w",stdout);    scanf("%d%d",&n,&p);    for (int i=1;i<n;++i) scanf("%d%d",&x,&y),add(x,y),add(y,x);    for (int i=1;i<=n;++i) scanf("%d",&a[i]);    f[0]=999999999;    now_sz=n;    getrot(1,0);    work(rot);    ans+=n;    printf("%d",ans);    return 0;}

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