Hash小结
Poj1480Eqs
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 ->a1x13+ a2x23+ a3x3=-(a4x43+ a5x53
问有多少个满足等式的非零x1,x2,x3,x4,x5组。)
中途相遇法，枚举x1,x2,x3得到左边式子的值插入hash表，然后枚举x4,x5找到对应的值就行了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL ;
using namespace std;
const int maxn = 1e6 + 7;
const int Hash = 1e5 + 7;
struct HashMap
{
int head[Hash + 10]; int size; int next[maxn*2]; int val[maxn*2]; int cnt[maxn*2];
void init()
{
size = 0; memset(head, -1, sizeof(head));
}
int ask(int x)
{
int k = (x%Hash + Hash) % Hash;
for (int i = head[k]; i != -1; i = next[i]){
if (val[i] == x) return cnt[i];
}
return 0;
}
void insert(int x)
{
int k = (x%Hash + Hash) % Hash;
for (int i = head[k]; i != -1; i = next[i]){
if (val[i] == x){
cnt[i] ++; return;
}
}
val[size] = x; cnt[size] = 1; next[size] = head[k];
head[k] = size++;
return;
}
}m;
int gao(int i)
{
return i*i*i;
}
int main()
{
int a, b, c, d, e;
while (cin>>a>>b>>c>>d>>e){
int ans = 0;
m.init();
for (int i = -50; i <= 50; i++)if (i){
for (int j = -50; j <= 50; j++)if (j){
for (int k = -50; k <= 50; k++)if (k){
int t = a*gao(i) + b*gao(j)+c*gao(k);
m.insert(t);
}
}
}
for (int i = -50; i <= 50; i++)if (i){
for (int j = -50; j <= 50; j++)if (j){
int t = d*gao(i) + e*gao(j);
ans += m.ask(-t);
}
}
cout << ans << endl;
}
return 0;
}
Poj2549 Sumsets
题意：Given S, a set of integers, find the largest d such that a + b + c = d where a, b, c, and d are distinct elements of S.
还是用中途相遇法搞下就好了。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL;
using namespace std;
const int Hash = 1e6 + 10;
const int maxn = 1e6 + 10;
struct HashMap
{
int head[Hash + 10]; int size; int next[maxn + 10]; int val[maxn + 10]; int a[maxn + 10], b[maxn + 10];
void init()
{
size = 0; memset(head, -1, sizeof(head));
}
void insert(int t, int x, int y)
{
int k = (t%Hash + Hash) % Hash;
for (int i = head[k]; i != -1; i = next[i]){
if (val[i] == t){
if (a[i] == x&&b[i] == y) return;
}
}
val[size] = t; next[size] = head[k]; a[size] = x; b[size] = y; head[k] = size++;
}
int ask(int t, int x, int y)
{
int k = (t%Hash + Hash) % Hash;
for (int i = head[k]; i != -1; i = next[i]){
if (val[i] == t){
if (a[i] != x&&b[i] != x&&a[i] != y&&b[i] != y) return 1;
}
}
return 0;
}
}m;
int cmp(const int &a, const int &b)
{
return a > b;
}
int a[1111];
int main()
{
int n;
while (scanf("%d", &n) != EOF&&n){
for (int i = 0; i < n; i++){
scanf("%d", &a[i]);
}
m.init();
sort(a, a + n, cmp);
for (int i = 0; i < n; i++){
for (int j = 0; j < n; j++){
if (i == j)continue;
m.insert(a[i] + a[j], i, j);
}
}
int flag = 1; int pos;
for (int i = 0; flag&&i < n; i++){
for (int j = 0; flag&&j < n; j++){
if (i == j) continue;
int t = m.ask(a[i] - a[j], i, j);
if (t){
flag = 0; pos = i; break;
}
}
}
if (flag){
printf("no solution\n");
}
else printf("%d\n", a[pos]);
}
return 0;
}
Poj3274 Gold Balanced Lineup
就是给出一些数，找到最大的连续区间使得区间内对应的二进制位上的1的个数相等。
若区间l，r满足条件，sum[i][j] 表示前i个数的第j位上的1的个数和。
sum[r][1] - sum[l][1] = sum[r][2] - sum[l][2].....->sum[r][1] - sum[r][2] = sum[l][1] - sum[l][2];
每一位都减去某位的值过以后，若是存在l,r区间满足条件则处理过后的对应位的值相等，此时用hash搞下。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL;
using namespace std;
const int Hash = 1e6 + 10;
const int maxn = 2e5 + 10;
int k;
struct HashMap
{
int head[Hash + 10]; int next[maxn + 10]; int val[maxn + 10]; int ans[maxn + 10][31];
int pos[maxn + 10];
int size;
void init()
{
size = 0; memset(head, -1, sizeof(head));
}
int ask(int a[])
{
int t = 0;
for (int i = 0; i < k; i++) t += a[i];
int gg = (t%Hash + Hash) % Hash;
for (int i = head[gg]; i != -1; i = next[i]){
if (val[i] == t){
int flag = 0;
for (int j = 0; j < k; j++) {
if (ans[i][j] != a[j]) {
flag = 1; break;
}
}
if (flag == 0) return pos[i];
}
}
return -2;
}
void insert(int a[], int p)
{
int t = 0;
for (int i = 0; i < k; i++) t += a[i];
int gg = (t%Hash + Hash) % Hash;;
for (int i = head[gg]; i != -1; i = next[i]){
if (val[i] == t){
int flag = 0;
for (int j = 0; j < k; j++){
if (ans[i][j] != a[j]){
flag = 1; break;
}
}
if (flag == 0) return;
}
}
val[size] = t; next[size] = head[gg];
pos[size] = p;
for (int i = 0; i < k; i++) ans[size][i] = a[i];
head[gg] = size++;
}
}m;
int cnt[100];
void gao(int a)
{
int len = 0;
while (a){
cnt[len++] += a % 2;
a >>= 1;
}
}
int main()
{
int n; int a;
while (scanf("%d%d", &n, &k) != EOF){
int Max = 0;
memset(cnt, 0, sizeof(cnt));
m.init(); m.insert(cnt, -1);
for (int i = 0; i < n; i++){
scanf("%d", &a); gao(a);
for (int j = k - 1; j >= 0; j--) cnt[j] -= cnt[0];
int t = m.ask(cnt);
if (t != -2) Max = max(Max, i - t);
m.insert(cnt, i);
}
printf("%d\n", Max);
}
return 0;
}
Hdu3333 Turing Tree
区间长度3e4 ，1e5个询问。
给出区间，求区间内不同元素和的个数。
开始用莫队算法多了sqrt(n)的复杂度，超时了(或许姿势写戳了).后来搜了下题解，发现按照查询的右端点排序，然后线段树边插入，边查询就好了。
到第i个元素的时候，只保留i这个大小的数最后一次出现的位置，前面出现过的位置上的数置0。由于是按右端点排序，在到达第k条询问的右端点的时候，保证到了当前位置区间内所有重复的数只有一个。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL;
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const LL maxn = 33333;
LL sum[maxn << 2];
LL last[maxn];
LL n;
LL a[maxn];
LL b[maxn];
void build(LL l, LL r, LL rt)
{
sum[rt] = 0;
if (l == r) return;
LL mid = (l + r) >> 1;
build(lson); build(rson);
}
void up(LL rt)
{
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void update(LL pos, LL add, LL l, LL r, LL rt)
{
if (l == r){
sum[rt] = add; return;
}
LL mid = (l + r) >> 1;
if (pos <= mid) update(pos, add, lson);
else update(pos, add, rson);
up(rt);
}
LL ask(LL L, LL R, LL l, LL r, LL rt)
{
if (L <= l&&r <= R) return sum[rt];
LL mid = (l + r) >> 1;
LL ans = 0;
if (L <= mid) ans += ask(L, R, lson);
if (R>mid) ans += ask(L, R, rson);
return ans;
}
void init()
{
vector<LL> q1;
for (LL i = 0; i < n; i++) q1.push_back(a[i]);
sort(q1.begin(), q1.end());
q1.erase(unique(q1.begin(), q1.end()), q1.end());
for (LL i = 0; i < n; i++) b[i] = lower_bound(q1.begin(), q1.end(), a[i]) - q1.begin();
}
struct Node
{
LL l; LL r; LL id;
}node[111111];
LL cmp(const Node &a1, const Node &b1)
{
return a1.r < b1.r;
}
LL ans[111111];
int main()
{
LL T;
cin >> T;
LL q;
while (T--){
cin >> n;
build(0, n - 1, 1);
for (LL i = 0; i < n; i++){
scanf("%I64d", &a[i]);
}
init();
cin >> q;
for (LL i = 0; i < q; i++){
scanf("%I64d%I64d", &node[i].l, &node[i].r); node[i].l -= 1; node[i].r -= 1;
node[i].id = i;
}
memset(last, -1, sizeof(last));
sort(node, node + q, cmp);
LL pos = 0;
for (LL i = 0; i < q; i++){
while (pos <= node[i].r){
LL t = last[b[pos]];
if (t == -1){
update(pos, a[pos], 0, n - 1, 1);
last[b[pos]] = pos;
}
else{
update(last[b[pos]], 0, 0, n - 1, 1);
update(pos, a[pos], 0, n - 1, 1);
last[b[pos]] = pos;
}
pos++;
}
ans[node[i].id] = ask(node[i].l, node[i].r, 0, n - 1, 1);
}
for (LL i = 0; i < q; i++){
printf("%I64d\n", ans[i]);
}
}
return 0;
}
Poj3188
题意：把L个字母按顺序分成连续的B份，变成手机按键，然后给出字典，问这些字典用手机按键去按，有多少个单词不存在冲突。
暴力枚举划分方式，然后hash去判断冲突。在弄的时候因为用了map，超时了，调了好久。。写的很戳
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL;
using namespace std;
const int Hash = 128;
const int maxn = 1e6;
int askval(char *s)
{
int k = 0;
while (*s) k = (*s++) + (k << 6) + (k << 16) - k;
return (k & 0x7fffffff);
}
int B, L, n;
struct HashMap
{
int size; int head[Hash + 10]; int val[maxn + 10]; int next[maxn + 10]; char s[maxn][14];
int cnt[maxn + 10];
void init()
{
size = 0; memset(head, -1, sizeof(head));
}
int ask(char *s1)
{
int k = askval(s1);
int t = (k%Hash + Hash) % Hash;
for (int i = head[t]; i != -1; i = next[i]){
if (val[i] == k){
if (strcmp(s1, s[i]) == 0) return cnt[i];
}
}
return 0;
}
void insert(char *s1)
{
int k = askval(s1);
int t = (k%Hash + Hash) % Hash;
for (int i = head[t]; i != -1; i = next[i]){
if (val[i] == k) if (strcmp(s1, s[i]) == 0){
cnt[i]++; return;
}
}
val[size] = k; next[size] = head[t]; strcpy(s[size], s1); cnt[size] = 1;
head[t] = size++;
}
}T;
char str[1111][12];
int a[100];
int Max;
int p[100];
int m[30];
void gao(int pos, int sum)
{
if (sum == B){
T.init();
memset(m, 0, sizeof(m));
for (int i = 0; i<B - 1; i++){
for (int j = a[i]; j<a[i + 1]; j++) m[j] = i;
}
for (int i = a[B - 1]; i<L; i++) m[i] = B - 1;
int ans = 0;
for (int i = 0; i<n; i++){
char s1[14]; strcpy(s1, str[i]);
int len = strlen(s1);
for (int j = 0; j<len; j++) s1[j] = m[s1[j] - 'A'] + 'A';
int t = T.ask(s1);
if (!t) T.insert(s1), ans++;
if (t == 1){
ans--; T.insert(s1);
}
}
if (ans>Max){
Max = ans; for (int i = 0; i<B; i++) p[i] = a[i];
}
return;
}
if (sum == 0){
a[sum] = 0; gao(0, sum + 1);
}
else
for (int i = L - B + sum; i>pos; i--){
a[sum] = i;
gao(i, sum + 1);
}
}
int main()
{
while (scanf("%d%d", &B, &L) != EOF){
scanf("%d", &n);
a[0] = -1; p[0] = -1;
for (int i = 0; i < n; i++){
scanf("%s", str[i]);
}
Max = -1;
gao(0, 0);
printf("%d\n", Max);
for (int i = 0; i<B - 1; i++){
for (int j = p[i]; j<p[i + 1]; j++) printf("%c", j + 'A');
printf("\n");
}
for (int i = p[B - 1]; i<L; i++) printf("%c", i + 'A');
if (p[B - 1] <= L - 1) printf("\n");
}
return 0;
}