洛谷P1087 FBI树

STL大法好

二进制划分，然后用下标建图，然后后序遍历

get到如果要给string一位一位的赋值要先给string开好空间s.resize(10000)

代码

//By AcerMo
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm> 
using namespace std;
const int M=2500;
struct map
{
	string s;
	map(){s.resize(2500);}
}emm[M];
map getss(int k,int i)
{
	map a;int st=(i-1)*k;
	for (int j=0;j<k;j++)
	a.s[j]=emm[0].s[j+st];
	return a;
} 
vector<int>v[M];
void built(int x)
{
	if (x==1) return ;
	for (int i=x/2;i<x;i+=2)
	v[i/2].push_back(i),v[i/2].push_back(i+1);
	built(x/2);
}
void coutt(int x)
{
	int e=0,ans=0;
	while (emm[x].s[e]>='0') ans+=(emm[x].s[e++]-'0');
	if (ans==0) cout<<"B";
	else if (ans==e) cout<<"I";
	else cout<<"F";
	return ;
}
void dfs(int x)
{
	if (v[x].size()==0) return (void)(coutt(x));
	for (int i=0;i<v[x].size();i++) dfs(v[x][i]);
	return (void)(coutt(x));
}//后序遍历 
int main()
{
	int n,cnt=0;cin>>n;
	cin>>emm[0].s;n=pow(2,n);
	for (int k=1;k<=n;k*=2)
	{
		int len=n/k;
		for (int i=1;i<=k;i++)
			emm[++cnt]=getss(len,i);
	}//取二进制份划分 
	built(n*2);//建树	
	dfs(1);
	return 0;
}