USACO holstein, hamming

微博: http://t.sina.com.cn/g7tianyi

豆瓣：http://www.douban.com/people/Jackierasy/

广度优先搜索的好题，广搜的好处之一就是往往可以直接找到最短路径，而且，就本题而言，还可以大规模剪枝。

优化的话估计还可以用位运算，那样最后一组数据应该也可以在0s内AC掉。

 

/*ID: fairyroadTASK:holsteinLANG:C++*///#define DEBUG#include<deque>#include<fstream>#include<iostream>using namespace std;ifstream fin("holstein.in");ofstream fout("holstein.out");typedef deque<int> Queue;int V, G;int vitamin[25];int fertilizer[15][25];struct node{Queue currsum, path;node() : currsum(0), path(0) {}node(const deque<int>& cs, const deque<int>& p) : currsum(cs), path(p) {}};#ifdef DEBUGinline void printnode(const node& nd){    cout<<"currsum :  ";    for(size_t i = 0; i < nd.currsum.size(); ++i) cout<<nd.currsum[i]<<' ';    cout<<"\npath :  ";    for(size_t i = 0; i < nd.path.size(); ++i) cout<<nd.path[i]<<' ';    cout<<"\n";}#endifinline bool check(const Queue& sum){for(int i = 0; i < V; ++i)if(sum[i] < vitamin[i]) return false;return true;}inline void addi(Queue& sum, int index){    for(size_t i = 0; i < sum.size(); ++i)        sum[i]+=fertilizer[index][i];}int main(){fin>>V;for(int i = 0; i < V; ++i) fin>>vitamin[i];fin>>G;for(int i = 0; i < G; ++i)for(int j = 0; j < V; ++j)fin>>fertilizer[i][j];deque<node> Q;Queue start_sum(V,0), start_path;//start_sum.push_back(0);start_path.push_back(-1); // for initilization, don't fell wierdQ.push_back(node(start_sum, start_path));while(!Q.empty()){        node n = Q.front();        #ifdef DEBUG            printnode(n);        #endif        for(int i = n.path.back()+1; i < G; ++i)        {            Queue tmpsum(n.currsum), tmppath(n.path);            addi(tmpsum, i);            tmppath.push_back(i);            if(check(tmpsum))            {                fout<<tmppath.size()-1<<' ';                for(size_t j = 1; j < tmppath.size()-1; ++j)                    fout<<tmppath[j]+1<<' ';                fout<<i+1 <<"\n";                goto DONE;            }            Q.push_back(node(tmpsum, tmppath));        }        Q.pop_front();}DONE:return 0;}

 

 

hamming是水题，USACO对位运算的考察还是很全面的。但是位运算是基础。

 

/*ID: fairyroadTASK:hammingLANG:C++*/#include<fstream>using namespace std;ifstream fin("hamming.in");ofstream fout("hamming.out");int N, B, D;int res[64];// return the number of 1 in the binary representation of the xor of a and binline int counthd(int a, int b){    int count = 0, n = a^b;    while(n)    {        n &= (n-1);        ++count;    }    return count;}int main(){    fin>>N>>B>>D;    int num = 1;    for(int i = 1; num <= N && i <= (1<<B)-1; ++i)    {        bool flag = true;        for(int j = 0; j < num; ++j)            if(counthd(res[j], i) < D)            {                flag = false;                break;            }        if(flag) res[num++] = i;    }    // output    for (int i=0; i<N-1; i++) fout<<res[i]<<( ((i+1)%10==0)?"\n":" ");    fout<<res[N-1]<<endl;    return 0;}