POJ 1840 Eqs(枚举+HASH)

//枚举 + HASH
//第一个优化，先预处理立方，先打个表
//第二个优化，拆分枚举，先枚举X1,X2,再枚举X3,X4,X5，这样将复杂度降为O(100^3)
//第三个优化，搜索剪枝,枚举X1,X2的时候记录结果的最大最小值，当在枚举X3,X4,X5的时候如果超过这个范围则跳出
//在判重的时候使用HASH，简单的取模HASH函数即可
//处理冲突采用拉链法
//344MS
#include<iostream>
#define MAXHASH 916549
#define NEG 12500000
using namespace std;
int CUBE[101];
struct Result
{
int cnt;
int data;
}res[10005];
int head[MAXHASH],next[10005];
int m = 0;
void init_cube()
{
for(int i = -50;i <= 50;++i)
CUBE[i + 50] = i * i * i;
}
inline int cube(int x)
{
return CUBE[x + 50];
}
inline int hash(int x)
{
return x % MAXHASH;
}
void insert(int x)//插入哈希表头
{
int h = hash(x);
bool Find = 0;
for(int e = head[h];e != -1;e = next[e])
{
if(res[e].data == x)
{
++res[e].cnt;
Find = 1;
break;
}
}
if(!Find)
{
res[m].cnt = 1;
res[m].data = x;
next[m] = head[h];
head[h] = m++;
}
}
int serch(int x)//搜索
{
int h = hash(x);
for(int e = head[h];e != -1;e = next[e])
if(res[e].data == x)
return res[e].cnt;
return 0;
}
int main()
{
int a1,a2,a3,a4,a5;
int ans = 0;
int _max = -2147483647,_min = 2147483647;
memset(res,0,sizeof(res));
memset(head,-1,sizeof(head));
init_cube();
scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
for(int x1 = -50;x1 <= 50;++x1)
{
if(x1 == 0)	continue;
for(int x2 = -50;x2 <= 50;++x2)
{
if(x2 == 0)	continue;
int sum = a1 * cube(x1) + a2 * cube(x2) + NEG;
if(sum > _max)	_max = sum;
if(sum < _min)	_min = sum;
insert(sum);
}
}
for(int x3 = -50;x3 <= 50;++x3)
{
if(x3 == 0)	continue;
for(int x4 = -50;x4 <= 50;++x4)
{
if(x4 == 0)	continue;
for(int x5 = -50;x5 <= 50;++x5)
{
if(x5 == 0)	continue;
int sum = -a3 * cube(x3) + -a4 * cube(x4) + -a5 * cube(x5) + NEG;
if(sum < _min || sum > _max)	continue;//一个简单的搜索剪枝
ans += serch(sum);
}
}
}
printf("%d/n",ans);
return 0;
}