[usaco]2.2 Subset Sums

大概的题意是： 给定一个序列 1 。。N ，假设全集为U那么存在多少种情况 ： 两个子集A B其中A∩B=∅ ，A∪B=U， A元素的和== B元素的和。开始写了个递归，枚举，提交超时：

void work(int deep,int start){    if(deep >= n)    return;    if(ans > (sum -ans)) return ;    if(ans == sum-ans)    {        gcount ++;        /*FOR_1(i,1,n){            if(used[i] == 1)            fout<<i<< " ";         }        fout<<endl;*/        return ;    }     FOR_1(i,start + 1,n){        if(used[i] == 0){            used[i] = 1;            ans += i;            work(deep + 1,i);            used[i] = 0;            ans -= i;        }    } }

之后改用另外一种思路，f(n,ans) = f(n-1,ans-n) + f(n-1,ans) ,这个有点分治的思路，把问题的规模逐渐缩小，从而得到解。中间加了个优化，用一个二维数组存下中间计算的结果，这样就不会超时。

/*ID:fuxiang2PROG: subsetLANG: C++*/#include <iostream>#include <fstream>#include <stack>#include <string>#include <vector>#include <queue>#include <map>#include <list>#include <algorithm>#include <set>#include <cmath>#include <cstring>#include <cstdlib> #define REP(i, n) for (int i=0;i<int(n);++i)#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)#define REP_1(i, n) for (int i=1;i<=int(n);++i)#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)#define EACH(it, A) for (typeof(A.begin()) it=A.begin(); it != A.end(); ++it) using namespace std;ofstream fout ("subset.out");ifstream fin ("subset.in"); const int N = 40;int used[N];int n ;int sum;int ans ;int gcount ;int data[40][800];void work(int deep,int start){    if(deep >= n)    return;    if(ans > (sum -ans)) return ;    if(ans == sum-ans)    {        gcount ++;        /*FOR_1(i,1,n){            if(used[i] == 1)            fout<<i<< " ";         }        fout<<endl;*/        return ;    }     FOR_1(i,start + 1,n){        if(used[i] == 0){            used[i] = 1;            ans += i;            work(deep + 1,i);            used[i] = 0;            ans -= i;        }    } } int solve(int n ,int ans){    int msum = (n+1)*n/2;    if(ans <= 0) return 0;    if(msum < ans) return 0 ;    else if(msum == ans ) return 1 ;     if(data[n][ans] )        return data[n][ans] ;     data[n-1][ans-n] = solve(n-1,ans-n);    data[n-1][ans] = solve(n-1,ans);     return data[n-1][ans-n] + data[n-1][ans];}int main(){    fin>>n;    sum = (1+n)*n/2;    ans = 0;    if(sum%2 ==1){        fout<<"0"<<endl;    }    else {        //work(0,0);        //fout<<gcount/2 <<endl;        int re = solve(n,sum/2);        fout<<re<<endl;    }     //int re = solve(n,sum/2);    //fout<<re<<endl;    return 0; }

原始博客地址： http://www.fuxiang90.com/2012/07/usaco2-2-subset-sums/