Problem Arrangement ZOJ - 3777 (dp + pressure desired shape)

Original link: http://www.cnblogs.com/WTSRUVF/p/10800034.html

Happening - 3777 

An entry form is desired pressure dp

dp[i][j]

The current state is i, for the case when j fractional number 
then look at the code annotated

 

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <list>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#defineMEM (A, B) Memset (A, B, the sizeof (A))
 #define _ :: ios_base sync_with_stdio (0), cin.tie (0)
 // The freopen ( "1.txt", "R & lt", stdin) ; 
the using  namespace STD;
 const  int MAXN = 110000 , INF = 0x7FFFFFFF ;
 int n-, m;
 int STR [ 13 is ] [ 13 is ];
 int VIS [ 13 is ];
 int F [ 15 ];
 int DP [ . 1 << 12 is ] [ 550 ];    // current state of the case when the number is i, fraction J 


int GCD ( int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}


int main()
{
    f[0] = 1;
    for(int i = 1; i <= 12; i++)
        f[i] = f[i - 1] * i;

    int T;
    rd(T);
    while(T--)
    {
        mem(dp, 0);
        rd(n), rd(m);
        for(int i = 0; I <n-; I ++ ) 
        { 
            for ( int J = 0 ; J <n-; J ++ ) 
                RD (STR [I] [J]); 
        } 

        int CNT = 0 ; 
        DP [ 0 ] [ 0 ] = . 1 ;
         for ( int I = 0 ; I <( . 1 << n-); I ++)    // traversing all states 
        { 
            CNT = 0 ;
             for ( int J = 0 ; J <n-; J ++)   //Calculating a current has been selected before the number of 
                IF (I & ( . 1 << J)) CNT ++ ;
             for ( int J = 0 ; J <n-; J ++)   // Current cnt + 1 th position in which one should 
            {
                 IF (I & ( . 1 << J)) Continue ;
                 for ( int K = 0 ; K <= m; K ++) // because we do not know the specific position of the first few so do not know the specific numerical values can iterate over all (m The maximum was 500) 
                {
                     IF (K + STR [CNT] [J]> = m) DP [I | ( . 1 << J)] [m] = + DP [I] [K];
                     the else DP [I | (1 << j)][k + str[cnt][j]] += dp[i][k];
                }
            }
        }
        if(dp[(1 << n) - 1][m] == 0)
            printf("No solution\n");
        else
        {

            int d = gcd(dp[(1 << n) - 1][m], f[n]);
            printf("%d/%d\n", f[n]/d, dp[(1 << n) - 1][m]/d);

        }

    }

    return 0;
}

 

Reproduced in: https: //www.cnblogs.com/WTSRUVF/p/10800034.html