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 12500000using 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;}