pku 1840 Eqs 哈希处理

http://poj.org/problem?id=1840

刚拿到这道题时，想到的肯定是暴力枚举啊。肯定超时。想到了立方乘系数，如果再枚举的话，还是会超时的。。无语。。

最后看了看了结题报告。。果然霸气。。。思维还是不够灵活是。。

将式子a1*x1^3 +a2*x2^3 + a3 * x3^3 + a4 * x4^3 + a5 * x5^3 = 0;转化成 a1*x1^3 +a2*x2^3 + a3 * x3^3 = - a4 * x4^3 - a5 * x5^3 

两边相等，将左边的数利用hash建立映射，然后枚举右边的数和利用hash查找。。。

开散列：

挂链 891ms

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxn 107
#define N 9997
using namespace std;

struct node
{
    int num;
    node *next;
}*tagp[N],H[99999997];

int pos;
int pw[maxn];

int main()
{
    int i,j,k,p;
    int a1,a2,a3,a4,a5;
    for (i = 0; i < 101; ++i)
    {
        pw[i] = (i - 50)*(i - 50)*(i - 50);
    }
    while (~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))
    {
        pos = 0;
        memset(tagp,0,sizeof(tagp));
        for (i = 0; i < 101; ++i)
        {
            if (i == 50) continue;
            for (j = 0; j < 101; ++j)
            {
                if (j == 50) continue;
                for (k = 0; k < 101; ++k)
                {
                    if (k == 50) continue;
                    int tmp = a1*pw[i] + a2*pw[j] + a3*pw[k];
                    p = tmp%N;
                    if (p < 0) p = -p;
                    node *t = &H[pos++];
                    t->num = tmp;
                    t->next = tagp[p];
                    tagp[p] = t;
                }
            }
        }
        int count = 0;
        for (i = 0; i < 101; ++i)
        {
            if (i == 50) continue;
            for (j = 0; j < 101; ++j)
            {
                if (j == 50) continue;
                int tmp = -a4*pw[i] - a5*pw[j];
                p = tmp%N;
                if (p < 0) p = -p;
                node *q;
                for (q = tagp[p]; q != NULL; q = q->next)
                {
                    if (q->num == tmp) count++;
                }
            }
        }
        printf("%d\n",count);
    }
    return 0;
}

vector 2610MS

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#define maxn 107
#define N 9997
using namespace std;

int pw[maxn];
vector<int>hash[N];
int main()
{
    int i,j,k,p;
    int a1,a2,a3,a4,a5;
    for (i = 0; i < 101; ++i)
    {
        pw[i] = (i - 50)*(i - 50)*(i - 50);
    }
    while (~scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5))
    {
        for (i = 0; i < N; ++i) hash[i].clear();
        for (i = 0; i < 101; ++i)
        {
            if (i == 50) continue;
            for (j = 0; j < 101; ++j)
            {
                if (j == 50) continue;
                for (k = 0; k < 101; ++k)
                {
                    if (k == 50) continue;
                    int tmp = a1*pw[i] + a2*pw[j] + a3*pw[k];
                    p = tmp%N;
                    if (p < 0) p = -p;
                    hash[p].push_back(tmp);
                }
            }
        }
        int count = 0;
        for (i = 0; i < 101; ++i)
        {
            if (i == 50) continue;
            for (j = 0; j < 101; ++j)
            {
                if (j == 50) continue;
                int tmp = -a4*pw[i] - a5*pw[j];
                p = tmp%N;
                if (p < 0) p = -p;
                for (k = 0; k < hash[p].size(); ++k)
                {
                    if (hash[p][k] == tmp) count++;
                }
            }
        }
        printf("%d\n",count);
    }
    return 0;
}