TopCoder SRM 613 Div.2 C DFS+剪枝

昨天刚做了一个搜索题，感觉和这个的剪枝方法很像。

当把剩下的数都选上时依然满足条件，则用组合数优化。

sta[i] 标记的为 从第 i 个到 第 n 张牌一共出现多少个数，最多五十个数，状压就好了。

STL 真心不会调试。。

#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <stack>

#pragma comment(linker, "/STACK:1024000000");
#define EPS (1e-6)
#define LL long long
#define ULL unsigned long long int
#define _LL __int64
#define _INF 0x3f3f3f3f
#define Mod 1000000007

using namespace std;

int fn[51],sn[51];

LL sta[51];

LL ANS;

LL cal[51][51];

void Cal(int n)
{
    int i,j;
    cal[0][0] = 0;

    for(i = 1;i <= n; ++i)
    {
        for(j = 0;j <= i; ++j)
        {
            if(j == 1)
                cal[j][i] = i;
            else if(j == 0 || i == j)
                cal[j][i] = 1;
            else
                cal[j][i] = cal[j-1][i-1] + cal[j][i-1];
        }
    }
}

void dfs(int s,int len,int k,int ans,LL st)
{
    cout<<"s = "<<s<<" len = "<<len<<" k = "<<k<<" ans = "<<ans<<endl;

    if(ans > k)
        return ;

    if(s == len)
    {
        ANS++;
        return ;
    }

    int i,temp = 0;

    for(i = 1;i <= max(10,len); ++i)
    {
        if((st&(((LL)1)<<i)) == 0 && (sta[s]&(((LL)1)<<i)))
            temp++;
    }

    if(temp+ans <= k)
    {
        for(i = len-s;i >= 0; --i)
            ANS += cal[i][len-s];
        return ;
    }

    dfs(s+1,len,k,ans,st);

    if((st&(((LL)1) << fn[s])) == 0)
    {
        st += (((LL)1) << fn[s]);
        ans++;
    }

    if((st&(((LL)1) << sn[s])) == 0)
    {
        st += (((LL)1) << sn[s]);
        ans++;
    }

    dfs(s+1,len,k,ans,st);
}

class TaroCards
{
public :
    long long int getNumber(vector<int> f,vector<int> s,int k)
    {
        int len,i;

        for(len = f.size(),i = 0; i < len; ++i)
        {
            fn[i] = f[i];
            sn[i] = s[i];
        }

        sta[len-1] = (((LL)1) << fn[len-1]);

        if(fn[len-1] != sn[len-1])
        {
            sta[len-1] += (((LL)1) << sn[len-1]);
        }

        for(i = len-2; i >= 0; --i)
        {
            sta[i] = sta[i+1];

            if((sta[i]&(((LL)1) << fn[i])) == 0)
            {
                sta[i] += (((LL)1) << fn[i]);
            }

            if((sta[i]&(((LL)1) << sn[i])) == 0)
            {
                sta[i] += (((LL)1) << sn[i]);
            }
        }

        ANS = 0;

        Cal(len);

        dfs(0,len,k,0,0);

        return ANS;
    }

    friend void dfs(int s,int len,int k,int ans,LL st);
    friend void Cal(int n);
};