USACO Healthy Holsteins DFS

使用排列组合，遍历所有可能的情况C(1)+C(2)+C(3)……C(n)= 2^G种组合

数据规模不大，暴力过去最多也就是2^15 = 23768种情况

所以就暴力咯，不过还是Debug了一会

 

Source Code:

/*
ID: wushuai2
PROG: holstein
LANG: C++
*/
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int M = 660000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;
const int MAX_N = 20         ;
const int MAXSIZE = 101000000;

int vv[30], aa[20][30];
int ans[30], a[30], cur[30];
int v, g;

void dfs(int n, int w, int a[]){

    int i, j;
    memset(cur, 0, sizeof(cur));
    for(i = 1; i <= a[0]; ++i){
        for(int j = 0; j < v; ++j){
            cur[j] += aa[a[i]][j];
        }
    }
    for(i = 0; i < v; ++i){
        if(cur[i] < vv[i]) break;
    }
    if(i == v){
        if(w < ans[0]){
            /*
            for(int i = 1; i <= a[0]; ++i){
                cout << a[i] + 1 << ' ';
            }
            cout << endl;
            */
            for(i = 0; i <= a[0]; ++i){
                ans[i] = a[i];
            }
            return;
        }
    }
    for(int i = n + 1; i < g; ++i){
        ++a[0];
        a[a[0]] = i;
        dfs(i, w + 1, a);
        --a[0];
    }
}

int main() {
    ofstream fout ("holstein.out");
    ifstream fin ("holstein.in");
    int i, j, k, t, n, s, c, w, q;

    fin >> v;
    for(i = 0; i < v; ++i){
        fin >> vv[i];
    }
    fin >> g;
    for(i = 0; i < g; ++i){
        for(j = 0; j < v; ++j){
            fin >> aa[i][j];
        }
    }
    ans[0] = INF;
    for(i = 0; i < g; ++i){
        a[0] = 1;
        a[1] = i;
        dfs(i, 1, a);
    }
    fout << ans[0];
    for(i = 1; i <= ans[0]; ++i){
        fout << ' ' << ans[i] + 1;
    }
    fout << endl;

    fin.close();
    fout.close();
    return 0;
}