Topcoder SRM 148 Div2 1000(dfs搜索+hash判重)

Problem Statement

 

9 numbers need to be arranged in a magic number square. A magic number square is a square of numbers that is arranged such that every row and column has the same sum. For example:

1 2 33 2 12 2 2

Create a class MNS containing a method combos which takes as an argument a vector <int>numbers and returns the number of distinct ways those numbers can be arranged in a magic number square. Two magic number squares are distinct if they differ in value at one or more positions. For example, there is only one magic number square that can be made of 9 instances of the same number.

Definition

 Class:MNSMethod:combosParameters:vector <int>Returns:intMethod signature:int combos(vector <int> numbers)(be sure your method is public)

Limits

 Time limit (s):2.000Memory limit (MB):64

Notes

-Unlike some versions of the magic number square, the numbers do not have to be unique.

Constraints

-numbers will contain exactly 9 elements.-each element of numbers will be between 0 and 9, inclusive.

Examples

0)  

{1,2,3,3,2,1,2,2,2}

Returns: 18

 1)  

{4,4,4,4,4,4,4,4,4}

Returns: 1

 2)  

{1,5,1,2,5,6,2,3,2}

Returns: 36

 3)  

{1,2,6,6,6,4,2,6,4}

Returns: 0

dfs搜索。用string存储图形，map<string,bool>判重。

能省就省。

这种题应该从想到写15分钟内搞定。太慢啦。

//Hello. I'm Peter.#include<cstdio>#include<iostream>#include<sstream>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<algorithm>#include<functional>#include<cctype>#include<ctime>#include<stack>#include<queue>#include<vector>#include<set>#include<map>using namespace std;typedef long long ll;typedef long double ld;#define peter cout<<"i am peter"<<endl#define input freopen("data.txt","r",stdin)#define randin srand((unsigned int)time(NULL))#define INT (0x3f3f3f3f)*2#define LL (0x3f3f3f3f3f3f3f3f)*2#define gsize(a) (int)a.size()#define len(a) (int)strlen(a)#define slen(s) (int)s.length()#define pb(a) push_back(a)#define clr(a) memset(a,0,sizeof(a))#define clr_minus1(a) memset(a,-1,sizeof(a))#define clr_INT(a) memset(a,INT,sizeof(a))#define clr_true(a) memset(a,true,sizeof(a))#define clr_false(a) memset(a,false,sizeof(a))#define clr_queue(q) while(!q.empty()) q.pop()#define clr_stack(s) while(!s.empty()) s.pop()#define rep(i, a, b) for (int i = a; i < b; i++)#define dep(i, a, b) for (int i = a; i > b; i--)#define repin(i, a, b) for (int i = a; i <= b; i++)#define depin(i, a, b) for (int i = a; i >= b; i--)#define pi 3.1415926535898#define eps 1e-6#define MOD 1000000007#define MAXN 100100#define N#define Mint ans;map<string,bool>app;void dfs(vector<int>ma[3],vector<int>belong,vector<int>d){    int len=gsize(belong);    if(len==0)    {        int t1,t2,t3,t4,t5,t6,t7,t8,t9,sum1;        t1=ma[0][0],t2=ma[0][1],t3=ma[0][2];        t4=ma[1][0],t5=ma[1][1],t6=ma[1][2];        t7=ma[2][0],t8=ma[2][1],t9=ma[2][2];        string s;        s.clear();        s+=d[t1]+'0';s+=d[t2]+'0';s+=d[t3]+'0';        s+=d[t4]+'0';s+=d[t5]+'0';s+=d[t6]+'0';        s+=d[t7]+'0';s+=d[t8]+'0';s+=d[t9]+'0';        if(!app[s]) app[s]=true;        else return;        sum1=d[t1]+d[t2]+d[t3];        int sum2;        sum2=d[t4]+d[t5]+d[t6];        if(sum1!=sum2) return;        sum2=d[t7]+d[t8]+d[t9];        if(sum1!=sum2) return;        sum2=d[t1]+d[t4]+d[t7];        if(sum1!=sum2) return;        sum2=d[t2]+d[t5]+d[t8];        if(sum1!=sum2) return;        sum2=d[t3]+d[t6]+d[t9];        if(sum1!=sum2) return;        ans++;        return;    }    int r=0;    if(len==9) r=0;    else if(len==6) r=1;    else if(len==3) r=2;    rep(i,0,len)    {        rep(j,0,len)        {            if(j==i ) continue;            rep(k,0,len)            {                if(k==i || k==j) continue;                vector<int>nextbelong;                nextbelong.clear();                if(len>3)                {                    rep(h,0,len)                    {                        if(h==i || h==j || h==k) continue;                        nextbelong.pb(belong[h]);                    }                }                ma[r].clear();                ma[r].pb(belong[i]);                ma[r].pb(belong[j]);                ma[r].pb(belong[k]);                dfs(ma,nextbelong,d);            }        }    }}class MNS{public:    int combos(vector <int> numbers)    {        ans=0;        vector<int>ma[3],belong;        rep(i,0,3)        {            ma[i].clear();        }        belong.clear();        rep(i,0,9)        {            belong.pb(i);        }        app.clear();        dfs(ma,belong,numbers);        return ans;    }};