POJ 1186 hash 线性探测再散列

一直以来我都觉得线性探测再散列这个方法一定很低效

结果碰到今天这题之后我惊奇的发现 链表居然被线性探测再散列秒杀了

这个hash方法是将大数取模，放到一个位置上，如果这个位置被占用了，就往后移1格，再被占再移动知道能 放到某个位置上。详细见数据结构书

真是太神奇了

懒得写DFS  直接for的

所以代码看起来比较多，但是也比较直观了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#define eps 1e-5
#define MAXN 4000037
#define MAXM 7300037
#define INF 1000000000
using namespace std;
int mod = 4000037;
int n, m;
int vis[MAXN], cnt[MAXN], hash[MAXN];
int locate(int key)
{int pos;pos = (key % mod + mod) % mod;while(vis[pos] && hash[pos] != key)if (++pos >= mod) pos -= mod;return pos;
}
void add(int key)
{int pos = locate(key);cnt[pos] ++;vis[pos] = 1;hash[pos] = key;
}
int f[11], p[11];
int numpow(int x, int num)
{int ans = 1;for(int i = 1; i <= num; i++) ans *= x;return ans;
}
int num[7][155];
int main()
{scanf("%d", &n);scanf("%d", &m);for(int i = 1; i <= n; i++) scanf("%d%d", &f[i], &p[i]);for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)num[i][j] = f[i] * numpow(j, p[i]);int ans = 0;if(n == 1){for(int i = 1; i <= m; i++)if(num[1][i] == 0) ans++;}else if(n == 2){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)if(num[1][i] + num[2][j] == 0) ans++;}else if(n == 3){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)for(int k = 1; k <= m; k++)if(num[1][i] + num[2][j] + num[3][k] == 0) ans++;}else if(n == 4){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)add(-(num[1][i] + num[2][j]));for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)ans += cnt[locate(num[3][i] + num[4][j])];}else if(n == 5){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)add(-(num[1][i] + num[2][j]));for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)for(int k = 1; k <= m; k++)ans += cnt[locate(num[3][i] + num[4][j] + num[5][k])];}else if(n == 6){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)for(int k = 1; k <= m; k++)add(-(num[1][i] + num[2][j] + num[3][k]));for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)for(int k = 1; k <= m; k++)ans += cnt[locate(num[4][i] + num[5][j] + num[6][k])];}printf("%d\n", ans);return 0;
}