[usaco]2.3 Zero Sum

1 2 3 4 5 6 7 ，在里面填 + - ，不填就数字合在一起， 比如2 3 = 23 了， 数字的顺序是一定的，那么数字之间的符号 就可以看作是不同的状态，dfs 遍历所有的情况。

/*
ID:fuxiang2
PROG: zerosum
LANG: 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 ("zerosum.out");
ifstream fin ("zerosum.in");

const int N = 10;
int flag
;
int n;
int ans ;

int hash[1100000];
int compute()
{
int num = 0;
FOR_1(i,1,n)
num = num *4 + flag[i];
return num;
}
void dfs(int deep)
{
if(deep == n  ){
flag[0] = 2;
int num = 0;
int t = 1;
int pre = 0;
FOR_1(i,1,n){

if(flag[i-1] ==0 )
return ;
if(flag[i-1] == 1)
t = t*10 + i;
else if(flag[i-1] == 2){
if(pre == 2)
num += t;
else if(pre == 3)
num -= t;
t = i;
pre = 2;
}
else if(flag[i-1] == 3){
if(pre ==2)
num += t;
else if(pre == 3)
num -= t;
t = i;
pre = 3;
}
}
if(pre == 2)
num += t;
else if(pre ==3)
num -= t;

if(num ==0 ){
fout<<1;
FOR_1(i,2,n){
if(flag[i-1] == 1)
fout<< " ";
else if(flag[i-1] == 2)
fout<< "+";
else if(flag[i-1] == 3)
fout<<"-";
fout<<i;

}
fout << endl;
}
}// end if

FOR_1(i,1,3){
flag[deep] = i;
int num = compute();
if(hash[num] == 0){
hash[num] = 1;
dfs(deep + 1);
}
flag[deep] = 0;

}

}
int main()
{
fin>>n;

dfs(1);
return 0;
}

原始博客地址 ： http://www.fuxiang90.com/?p=1136