topcoder13185 TreePuzzle

https://community.topcoder.com/stat?c=problem_statement&pm=13185

被wck屠了。

考试时候想分类讨论，结果发现情况有点复杂，最后还是没调出来。

回去看了看题解，发现好像是树形DP，状态记得很巧妙。

假设当前红点在$x$，从$fa$来，容易知道此时$fa$是空的。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj

using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef complex<DB> CP;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define fill(a,l,r,v) fill(a+l,a+r+1,v)
#define re(i,a,b)  for(i=(a);i<=(b);i++)
#define red(i,a,b) for(i=(a);i>=(b);i--)
#define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++)
#define fi first
#define se second
#define m_p(a,b) make_pair(a,b)
#define p_b(a) push_back(a)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))

template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

int gint()
{
int res=0;bool neg=0;char z;
for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
if(z==EOF)return 0;
if(z=='-'){neg=1;z=getchar();}
for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
return (neg)?-res:res;
}
LL gll()
{
LL res=0;bool neg=0;char z;
for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());
if(z==EOF)return 0;
if(z=='-'){neg=1;z=getchar();}
for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());
return (neg)?-res:res;
}

const int maxn=300;

int n;
int d[maxn+10],g[maxn+10][maxn+10];
int mark[maxn+10];
int sz[maxn+10][maxn+10];//sz[i][j]表示i的父亲为j时，子树i的大小
int bl[maxn+10];
int res[maxn+10];

void addedge(int u,int v){g[u][++d[u]]=v;}

int calsz(int x,int fa)
{
int &res=sz[x][fa],j;
if(res)return res;
res=1;
re(j,1,d[x])if(g[x][j]!=fa)res+=calsz(g[x][j],x);
return res;
}
int calbl(int x,int fa)
{
int &res=bl[x],j;
res=mark[x];
re(j,1,d[x])if(g[x][j]!=fa)res+=calbl(g[x][j],x);
return res;
}

queue<int>Q;
int vis[maxn+10][maxn+10][maxn+10];
void push(int x,int fa,int c)
{
if(vis[x][fa][c])return;
vis[x][fa][c]=1;
Q.push((x<<20)|(fa<<10)|c);
}

int main()
{
freopen("puzzle.in","r",stdin);
freopen("puzzle.out","w",stdout);
int i,j;
n=gint();
re(i,1,n){int fa=gint()+1;if(fa)addedge(fa,i),addedge(i,fa);}
re(i,1,n)re(j,1,d[i])calsz(i,g[i][j]);
re(i,1,n)mark[i]=gint();mark[1]=0;
calbl(1,-1);
int total=bl[1];
re(j,1,d[1])
{
int v=g[1][j];
if(sz[v][1]>bl[v])push(v,1,bl[v]);
}
res[1]=1;
while(!Q.empty())
{
int status=Q.front(),x=status>>20,fa=(status>>10)&1023,c=status&1023;Q.pop();//当前红点在x，从fa来，子树内的黑点数为c
res[x]=1;
if(total-c<sz[fa][x])push(fa,x,total-c);
int sc=0;
re(j,1,d[x])if(g[x][j]!=fa)sc+=sz[g[x][j]][x];
re(j,1,d[x])if(g[x][j]!=fa)
{
int v=g[x][j];
for(int c2=0;c2<=c && c2<sz[v][x];c2++)if(c-c2<=sc-sz[v][x])push(v,x,c2);
}
}
re(i,1,n)PF("%d ",res[i]);PF("\n");
return 0;
}

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