google codejam 2008 round2

Problem A. Cheating a Boolean Tree

模拟题，可将dfs的结果保存到数组进行优化

 #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <fstream>
#include <iomanip>
#include <cmath>
#include <string>
#include <string.h>
#include <sstream>
#include <cctype>
#include <climits>
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <vector>
#include <iterator>
#include <algorithm>
#include <stack>
#define _clr(x,y) memset(x,y,sizeof(x))
#define _inf(x) memset(x,0x3f,sizeof(x))
#define pb push_back
#define mp make_pair
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
typedef long long LL;
struct node
{
	int g,c,val;
}tree[10000+5];
int m;
int dfs(int x,int v)
{
	int res=INF,res2=INF;
	if(x>(m-1)/2) if(v==tree[x].val) return 0;else return INF;
	int l1,l0,r1,r0;
	l1=dfs(x*2,1);l0=dfs(x*2,0);r1=dfs(x*2+1,1);r0=dfs(x*2+1,0);
	if(tree[x].g==1)
	{
		if(v==1)
		{
			res=l1+r1;
		}
		else
		{
			res=min(l1+r0,r1+l0);
			res=min(res,r0+l0);
		}
	}
	else
	{
		if(v==0)
		{
			res=l0+r0;
		}
		else
		{
			res=min(l1+r0,r1+l0);
			res=min(res,r1+l1);
		}
	}
	if(tree[x].c==1)
	{
		if(tree[x].g==0)
		{
			if(v==1)
			{
				res2=l1+r1;
			}
			else
			{
				res2=min(l1+r0,r1+l0);
				res2=min(res2,r0+l0);
			}
		}
		else
		{
			if(v==0)
			{
				res2=l0+r0;
			}
			else
			{
				res2=min(l1+r0,r1+l0);
				res2=min(res2,r1+l1);
			}
		}
	}
	res=min(res,res2+1);
	if(res>=INF) res=INF;
	//cout<<x<<" "<<v<<" "<<res<<endl;
	return res;
}
int main()
{
	//ios_base::sync_with_stdio(false);cin.tie(0);
	#ifndef ONLINE_JUDGE
		//freopen("A-small-practice.in","r",stdin);
		//freopen("out.txt","w",stdout);
	#endif

	int T;
	cin>>T;
	for(int cas=1;cas<=T;cas++)
	{
		_clr(tree,0);
        int v;
        cin>>m>>v;
        for(int i=1;i<=(m-1)/2;i++)
        {
        	cin>>tree[i].g>>tree[i].c;
		}
		for(int i=(m-1)/2+1;i<=m;i++)
        {
        	cin>>tree[i].val;
		}
		int ans=dfs(1,v);
		cout<<"Case #"<<cas<<": ";
		if(ans==INF) cout<<"IMPOSSIBLE"<<endl;
		else cout<<ans<<endl;
	}

    return 0;
}