Hash的应用
学习资料:论文一,论文二
Rabin-Karp string search algorithm
1.pku-1200
描述:求在文本中出现的不同子串(给定长度)的个数。
分析:最初"You may assume that the maximum number of substrings formed by the possible set of characters does not exceed 16 Millions."
这句理解有误,正确的理解是:nc^n <= 16,000,000,根据这个条件可以确定用nc进制hash(R-K algorithm),并且不需要处理冲突。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <stdio.h>
#include <string.h>
#define NL 20000000
char s[NL];
int n, nc;
int b[30];
int v[255];
bool hash[NL];
int main() {
while (scanf("%d%d", &n, &nc) != EOF) {
scanf("%s", s);
b[0] = 1;
for (int i = 1; i < n; i++) {
b[i] = b[i - 1] * nc;
}
int len = strlen(s);
if (len < n) {
printf("0\n");
continue;
}
memset(v, -1, sizeof(v));
//提取出字符集,对应到0~nc-1
for (int i=0, j=0; i<len; i++) {
if (v[s[i]]<0) {
v[s[i]] = j++;
}
}
//R-K algorithm
memset(hash, 0, sizeof (hash));
int key = 0;
for (int i = 0; i < n; i++) {
key += b[i]*v[s[i]];
}
int sum = 1;
hash[key] = 1;
for (int i = 1; i <= len - n; i++) {
key = (key - v[s[i - 1]]) / nc + v[s[i + n - 1]] * b[n - 1];
if (!hash[key]) {
hash[key] = 1;
sum++;
}
}
printf("%d\n", sum);
}
return 0;
}
//79ms
2.pku-1635[zju-1990]
描述:判定树的同构(根结点固定),树的最小表示法。
反思:用C实现很麻烦,换成string,但效率就不是很高了,TLE一次。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
#define NL 3010
void srt(string s, int n, string &cs) {
vector<string> sub;
string ss;
int z, o, t = 0, i = 0, k = 0;
z = o = 0;
while (i < n) {
if (s[i] == '0') z++;
else o++;
k++;
/*
* 0和1的个数相同时说明已经遍历了结点的一个分支,去掉开头的0和结尾的1就是相应的子树;
* 然后递归,将所有的子树按字典序排列,得到最小表示法,最后比较是否相同。
*/
if (z == o) {
if (k > 2) {
srt(s.substr(t+1, k-2), k - 2, ss);
ss.insert(0, "0");
ss.insert(k-1, "1");
sub.push_back(ss);
}else {
sub.push_back("01");
}
t = i + 1;
k = 0;
z = 0;
o = 0;
}
i++;
}
sort(sub.begin(), sub.end());
cs = "";
vector<string>::iterator it = sub.begin();
while (it != sub.end()) {
cs += *it;
it++;
}
}
int main() {
// freopen("datain", "r", stdin);
int n;
string s1, s2, cs1, cs2;
cin >> n;
while (n--) {
cin >> s1 >> s2;
srt(s1, s1.length(), cs1);
srt(s2, s2.length(), cs2);
if (cs1 == cs2) {
cout << "same\n";
}else {
cout << "different\n";
}
}
return 0;
}
//469ms
3.poj-1971
描述:平面上n个点,能构成多少个平行四边形。
思路:根据定理“平行四边形的对角线相互平分”,求出C(n,2)条线段的中点,中点重合的线段可以组合构成平行四边形。(见下图)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#define EP 1e-10
#define NL 1001
using namespace std;
struct Node {
int x, y;
} p[NL];
int dcmp(double x, double y) {
if (fabs(x - y) < EP)
return 0;
return x < y ? -1 : 1;
}
struct L {
double mdx, mdy;
bool operator <(const L &a) const {
if (dcmp(mdx, a.mdx) == 0) {
return dcmp(mdy, a.mdy) < 0 ? 1 : 0;
}
return dcmp(mdx, a.mdx) < 0 ? 1 : 0;
}
} l[NL * NL];
int cmp(const void *a, const void *b) {
struct L *x = (struct L *) a;
struct L *y = (struct L *) b;
if (dcmp(x->mdx, y->mdx) == 0) {
return dcmp(x->mdy, y->mdy);
}
return dcmp(x->mdx, y->mdx);
}
int main() {
// freopen("data.in", "r", stdin);
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d%d", &p[i].x, &p[i].y);
}
int m = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++, m++) {
l[m].mdx = (p[i].x + p[j].x) * 1.0 / 2;
l[m].mdy = (p[i].y + p[j].y) * 1.0 / 2;
}
}
sort(l, l + m);
double px, py;
px = l[0].mdx;
py = l[0].mdy;
int oz = 0, sum = 0;
for (int i = 1; i < m; i++) {
if (fabs(px - l[i].mdx) < EP && fabs(py - l[i].mdy) < EP) {
oz++;
} else {
sum += (oz + 1) * oz / 2;
oz = 0;
px = l[i].mdx;
py = l[i].mdy;
}
}
sum += (oz + 1) * oz / 2;
printf("%d\n", sum);
}
return 0;
}
//1641ms
4.poj-2002
描述:平面上n个点,能构成多少个正方形。
思路:对点hash;枚举边,计算出对应的能与其构成正方形的点,用hash判断是否存在。
知识:已知两点(x1,y1) , (x2,y2) 对应的有向线段是(x2-x1,y2-y1), 与其垂直的有向线段可以表示为,(y2-y1,x1-x2) 或 (y1-y2, x2-x1)
ps: hash函数不同时间效率会有很大不同,需要优化
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <stdio.h>
#include <string.h>
#define NL 1001
#define MD 199997
#define ADD 20010
int hash[MD];
struct POINT {
int x, y;
}p[NL];
void dh(int k) {
int key = ((p[k].x+ADD)*1000+(p[k].y+ADD))%MD;
// int key = (p[k].x+p[k].y+MD+MD)%MD;
while (hash[key] >= 0) {
key = (key+1)%MD;
}
hash[key] = k;
}
int dh1(POINT po) {
int key = ((po.x+ADD)*1000+(po.y+ADD))%MD;
// int key = (po.x+po.y+MD+MD)%MD;
while (hash[key] >= 0) {
int t = hash[key];
if (p[t].x == po.x && p[t].y == po.y) {
return 1;
}
key = (key+1)%MD;
}
return 0;
}
int main()
{
// freopen("data.in", "r", stdin);
int n;
while (scanf("%d", &n) != EOF) {
if (!n) break;
memset(hash, -1, sizeof(hash));
for (int i=0; i<n; i++) {
scanf("%d%d", &p[i].x, &p[i].y);
dh(i);
}
int sum = 0;
POINT p1, p2, dr1, dr2;
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
dr1.x = p[i].y-p[j].y;
dr1.y = p[j].x-p[i].x;
dr2.x = p[j].y-p[i].y;
dr2.y = p[i].x-p[j].x;
p1.x = p[i].x+dr1.x;
p1.y = p[i].y+dr1.y;
p2.x = p[j].x+dr1.x;
p2.y = p[j].y+dr1.y;
int ok1, ok2;
ok1 = dh1(p1);
ok2 = dh1(p2);
if (ok1 & ok2) {
sum++;
}
p1.x = p[i].x+dr2.x;
p1.y = p[i].y+dr2.y;
p2.x = p[j].x+dr2.x;
p2.y = p[j].y+dr2.y;
ok1 = dh1(p1);
ok2 = dh1(p2);
if (ok1 & ok2) {
sum++;
}
}
}
printf("%d\n", sum/4);
}
return 0;
}
//1600+ms