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usaco PROB Checker Challenge 搜索

搜索的过程中用2进制保存状态,通过位运算判断可行性。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\\in1.txt","r",stdin);
    freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}
int res[3][20];
int buf[20];
int n;
int num;
int dfs(int k,int col,int dl,int dr)
{
    if(k>n)
    {
        if(num<3)
        {
            memcpy(res[num],buf,sizeof(buf));
        }
        num++;
        return 0;
    }
    for(int i=1;i<=n;i++)
    {
        if(
           !(col&(1<<i))&&
           !(dl&(1<<(k+i)))&&
           !(dr&(1<<(k-i+n)))
           )
        {
            buf[k]=i;
            dfs(k+1,col|(1<<i),dl|(1<<(k+i)),dr|(1<<(k-i+n)));
        }
    }
    return 0;
}
int main()
{
    freopen("checker.in","r",stdin);
    freopen("checker.out","w",stdout);

    scanf("%d",&n);
    num=0;
    dfs(1,0,0,0);
    for(int i=0;i<3;i++)
    {
        for(int j=1;j<=n;j++)
            printf("%d%c",res[i][j],j==n?'\n':' ');

    }
    printf("%d\n",num);
    return 0;
}
View Code

 

posted on 2014-02-20 19:18  BMan、  阅读(219)  评论(0编辑  收藏  举报

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