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考虑一种颜色对答案的贡献

把树中这种颜色的点都删掉,那么就会有很多的小树,这些小树中的点互相之间不会产生贡献,而不同树的两个点之间会产生贡献。

可以得到每一种颜色,点的sum值就是n - 所在小树的size

一个点的sum就是n * 颜色数 - 每种颜色节点所在小树的size

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 100011
int n,cnt=0,cs=0,head[N],col[N],siz[N],del[N],csz[N],tag[N];
ll ans[N],qwq=0;
struct edge{
    int to,nxt;
}e[N<<1];
il void add(int x,int y){
    e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;
}
bool v[N];
void dfs(int x,int f){
    siz[x]=1;
    int tt=del[col[f]];
    fl(i,x)if(to!=f){
        dfs(to,x);
        siz[x]+=siz[to];
    }
    ++del[col[x]];
    if(f){
        csz[x]=siz[x]-del[col[f]]+tt;//每种颜色节点所在小树的size
        del[col[f]]+=csz[x];
    }
}
void get(int x,int f){
    int oldtag=tag[col[f]];
    qwq+=csz[x]-tag[col[f]];
    tag[col[f]]=csz[x];
    ans[x]=1ll*n*cs-qwq+tag[col[x]];
    fl(i,x)if(to!=f)get(to,x);
    tag[col[f]]=oldtag;
    qwq-=csz[x]-tag[col[f]];
}
int main(){
    in(n);
    Fur(i,1,n){
        in(col[i]);
        if(!v[col[i]])v[col[i]]=1,++cs;
    }
    int x,y;
    Fur(i,1,n-1)
        in(x,y),
        add(x,y),add(y,x);

    dfs(1,0);
    Fur(i,1,100000)
    if(v[i]){
        qwq+=n-del[i];
        tag[i]=n-del[i];
    }
    get(1,0);
    Fur(i,1,n)out(ans[i],ln);
    flush();
}
LG 2664 树上游戏
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