模板汇总
模板汇总
也许是中学生涯最后一次打板子了呀...
KMP模式匹配
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1000005;
char a[N], b[N];
int a1, b1, Next[N];
int main(){
cin >> a + 1;
cin >> b + 1;
a1 = strlen(a + 1);
b1 = strlen(b + 1);
Next[1] = 0;
for(int i = 2, j = 0; i <= b1; i++){
while( j > 0 && b[j + 1] != b[i] ) j = Next[j];
if( b[i] == b[j + 1] ) j++;
Next[i] = j;
}
for(int i = 1, j = 0; i <= a1; i++){
while( j > 0 && a[i] != b[j + 1]) j = Next[j];
if(a[i] == b[j + 1]) j++;
if(j == b1) {
cout << i - j + 1 <<"\n";
j = Next[j];
}
}
for(int i = 1; i <= b1; i++) cout << Next[i] <<" ";
return 0;
}
Trie字典树
#include<iostream>
#include<cstdio>
using namespace std;
int Trie[N][27], end[p] = 1, tot = 1;
void insert(char* str){
int len = strlen(str), p = 1;
for(int k = 0; k < len; k++){
int ch = str[k] - 'a';
if(trie[p][ch] == 0) trie[p][ch] = ++tot;
p = trie[p][ch];
}
end[p] = 1;
}
bool search(char* str){
int len = strlen(str), p = 1;
for(int k = 0; k < len; k++){
p = trie[p][str[k]];
if(p == 0) return 0;
}
return end[p];
}
ST表
值得注意的是，log函数最好配强制类型转换（不然poj会编译错误），转换的时候要用括号把log全部括一下，初始化的时候k要加一。初始化的时候j从0开始，我被坑了无数次了
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int N = 100005;
int a[N][30], n, m, k;
int read(){
int x = 0, ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x;
}
int main(){
cin >> n >> m;
for(int i = 1; i <= n; i++) a[i][0] = read();
k = int(log(n) / log(2) + 1);
for(int i = 1; i <= k; i++)
for(int j = 0; j <= n - (1 << i) + 1; j++)
a[j][i] = max(a[j][i - 1], a[j + (1 << (i-1))][i - 1]);
while(m--){
int x = read(), y = read();
k = int(log(y - x + 1) / log(2));
cout << max(a[x][k], a[y - (1 << k) + 1][k]) << "\n";
}
return 0;
}
单源最短路径 \(dijkstra\)
注意打标记和判重都在取出队头的时候进行
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 500005;
int ver[N], head[N], Next[N], edge[N], tot;
int n, m, s;
void add(int x, int y, int z){
ver[++tot] = y, edge[tot] = z;
Next[tot] = head[x], head[x] = tot;
}
int read(){
int x = 0, ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x;
}
/*Dijkstra*/
int d[N], v[N];
void dijkstra(int s){
memset(d, 0x3f, sizeof d);
memset(v, 0, sizeof v);
priority_queue< pair< int, int > > q;
q.push(make_pair(0, s)); d[s] = 0;
while(q.size()){
int x = q.top().second; q.pop();
if(v[x]) continue;
v[x] = 1;
for(int i = head[x]; i; i = Next[i]){
int y = ver[i], z = edge[i];
if(d[y] > d[x] + z){
d[y] = d[x] + z;
q.push(make_pair(-d[y], y));
}
}
}
}
int main(){
cin >> n >> m >> s;
for(int i = 1; i <= m; i++){
int x = read(), y = read(), z = read();
add(x, y, z);
}
dijkstra(s);
for(int i = 1; i <= n; i++){
if(d[i] == 0x3f3f3f3f)
cout << "2147483647 ";
else cout << d[i] <<" ";
}
return 0;
}
单源最短路径\(SPFA\)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 500005;
int ver[N], head[N], Next[N], edge[N], tot;
int n, m, s;
void add(int x, int y, int z){
ver[++tot] = y, edge[tot] = z;
Next[tot] = head[x], head[x] = tot;
}
int read(){
int x = 0, ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x;
}
int d[N], v[N];
/* spfa */
void spfa(int s){
memset(d, 0x3f, sizeof d);
memset(v, 0, sizeof v);
queue<int> q;
q.push(s); v[s] = 1; d[s] = 0;
while(q.size()){
int x = q.front(); q.pop();
v[x] = 0;
for(int i = head[x]; i; i = Next[i]){
int y = ver[i], z = edge[i];
if(d[y] > d[x] + z){
d[y] = d[x] + z;
if(!v[x]) q.push(y), v[y] = 1;
}
}
}
}
int main(){
cin >> n >> m >> s;
for(int i = 1; i <= m; i++){
int x = read(), y = read(), z = read();
add(x, y, z);
}
spfa(s);
for(int i = 1; i <= n; i++){
if(d[i] == 0x3f3f3f3f)
cout << "2147483647 ";
else cout << d[i] <<" ";
}
return 0;
}
扩展欧几里得
\(x = \frac{c}{d}x_0 + k \frac{b}{d}, y = \frac {c}{d}y_0 - k \frac{a}{d}\)
#include<iostream>
#define ll long long
using namespace std;
int exgcd1(int a, int b, int &x, int &y){
if(b == 0) {
x = 1, y = 0;
return a;
}
int d = exgcd1(b, a % b, x, y);
int z = x; x = y, y = z - (a / b) * y;
return d;
}
ll exgcd(ll a, ll b, ll &x, ll &y){
if(b == 0){
x = 1, y = 0;
return a;
}
int d = exgcd(b, a % b, x, y);
int z = x; x = y, y = z - (a / b) * y;
return d;
}
int main(){
ll a, b, x, y, d;
cin >> a >> b;
d = exgcd(a, b, x, y);
cout << (x % (b / d) + (b / d)) % (b / d) <<"\n";
return 0;
}
最近公共祖先 LCA
- 树上倍增法
敲这个板子出了一堆错误。所以要记得空间开到两倍，lca的大于等于号。
错的太智障了
#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
const int N = 1000005;
int ver[N], head[N], Next[N], tot;
int n, m, s, k, f[N][25], d[N];
void add(int x, int y){
ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
}
void bfs(){
queue<int> q;
q.push(s); d[s] = 1;
while(q.size()){
int x = q.front(); q.pop();
for(int i = head[x]; i; i = Next[i]){
int y = ver[i];
if(d[y]) continue;
d[y] = d[x] + 1;
f[y][0] = x;
for(int j = 1; j <= k; j++)
f[y][j] = f[f[y][j - 1]][j - 1];
q.push(y);
}
}
}
int lca(int x, int y){
if(d[y] < d[x]) swap(x, y);
for(int i = k; i >= 0; i--)
if(d[f[y][i]] >= d[x]) y = f[y][i];
if(x == y) return x;
for(int i = k; i >= 0; i--)
if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
return f[x][0];
}
int read(){
int x = 0, ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x;
}
int main(){
cin >> n >> m >> s;
k = (int)( log(n) / log(2) ) + 1;
for(int i = 1; i < n; i++) {
int x = read(), y = read();
add(x, y), add(y, x);
}
bfs();
while(m--){
int x = read(), y = read();
cout << lca(x, y) <<"\n";
}
return 0;
}
- 树链剖分法
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 1000005;
int ver[N], head[N], Next[N], tot;
int d[N], son[N], f[N], top[N], v[N], size[N];
int n, m, s;
void add(int x, int y){
ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
}
void dfs1(int x){
v[x] = 1;
size[x] = 1;
for(int i = head[x]; i; i = Next[i]){
int y = ver[i];
if(v[y]) continue;
d[y] = d[x] + 1;
dfs1(y);
f[y] = x;
size[x] += size[y];
if(!son[x] || size[y] > size[son[x]])
son[x] = y;
}
}
void dfs2(int x, int p){
top[x] = p;
if(son[x]) dfs2(son[x], p);
for(int i = head[x]; i; i = Next[i]){
int y = ver[i];
if(y == f[x] || y == son[x]) continue;
dfs2(y, y);
}
}
int lca(int x, int y){
while(top[x] != top[y])
if(d[top[x]] > d[top[y]]) x = f[top[x]];
else y = f[top[y]];
if(d[x] > d[y]) return y;
return x;
}
int read(){
int x = 0, ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x;
}
int main(){
cin >> n >> m >> s;
for(int i = 1; i < n; i++) {
int x = read(), y = read();
add(x, y), add(y, x);
}
d[s] = 1;
dfs1(s);
dfs2(s, s);
while(m--){
int x = read(), y = read();
cout << lca(x, y) <<"\n";
}
return 0;
}
最小生成树 Kruskal
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 200005;
int n, m, f[N], ans = 0;
struct node{
int x, y, z;
}edge[N];
bool cmp(node a, node b){
return a.z < b.z;
}
int get(int x){
if(x == f[x]) return x;
return f[x] = get(f[x]);
}
int main(){
cin >> n >> m;
for(int i = 1; i <= n; i++) f[i] = i;
for(int i = 1; i <= m; i++){
cin >> edge[i].x >> edge[i].y >> edge[i].z;
}
sort(edge + 1, edge + m + 1, cmp);
for(int i = 1; i <= m; i++){
int x = edge[i].x, y = edge[i].y;
x = get(x), y = get(y);
if(x == y) continue;
f[x] = y;
ans += edge[i].z;
}
cout << ans;
return 0;
}
树的直径
void dp(int x){
v[x] = 1;
for(int i = head[x]; i; i = Next[i]){
int y = ver[i];
if(v[y]) continue;
dp(y);
ans = max(ans, d[x] + d[y] + edge[i]);
d[x] = max(d[x], d[y] + edge[i]);
}
}
树状数组
OI生涯学的第一个数据结构！
板子超好写，难在分析题目中能用前缀和、差分解决的地方。
void update(int x, int y){
for(; x <= n; x += x & -x)
s[x] += y;
}
int getsum(int x){
int ans = 0;
for(; x; x -= x & -x)
ans += s[x];
return ans;
}
归并排序
\(mid - i + 1\)背下来就好啦
#include<iostream>
using namespace std;
int a[100005], b[100005], ans;
void merge_sort(int l, int r){
if(l == r) return;
int mid = (l + r) >> 1;
merge_sort(l, mid);
merge_sort(mid + 1, r);
int i = l, j = mid + 1, k = l;
while(i <= mid || j <= r){
if(j > r || ( a[i] <= a[j] && i <= mid ))
b[k++] = a[i++];
else
b[k++] = a[j++], ans += mid - i + 1;
}
for(int i = l; i <= r; i++)
a[i] = b[i];
}
int main(){
int n;
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
merge_sort(1, n);
cout << ans;
return 0;
}
乘法逆元
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 3000005;
long long a[N], n, p;
int main(){
cin >> n >> p;
a[0] = a[1] = 1;
cout << 1 <<"\n";
for(int i = 2; i <= n; i++){
a[i] = (p - p / i) * a[p % i] % p;
cout << a[i] <<"\n";
}
return 0;
}
卢卡斯定理
#include<iostream>
#include<cstdio>
using namespace std;
const int N = 1000005;
long long a[N];
long long n, m, p;
long long fp(long long a, long long b){
long long ans = 1;
for(; b; b >>= 1){
if(b & 1) ans = ans * a % p;
a = a * a % p;
}
return ans;
}
long long C(long long n, long long m){
if(n < m) return 0;
return (a[n] * fp(a[m], p - 2)) % p * fp(a[n - m], p - 2) % p ;
}
long long Lucas(long long n, long long m){
if(m == 0) return 1;
return C(n % p, m % p) * Lucas(n / p, m / p) % p;
}
int read(){
int x = 0, ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x;
}
int main(){
// for(int i = 2; i < p; i++) a[i] = (p - p / i) * a[p % i] % p;
int t = read();
while(t--){
n = read(), m = read(), p = read();
a[0] = 1;
for(int i = 1; i <= p; i++) a[i] = a[i - 1] * i % p;
cout << Lucas(n + m, n) <<"\n";
}
return 0;
}
线段树
一个暑假的记忆，大概是我最有把握的模板了吧。
#include<iostream>
#include<cstdio>
using namespace std;
const int SIZE = 100005;
struct SegmentTree{
int l, r;
long long sum, add;
}t[SIZE * 4];
int a[SIZE];
void build(int p, int l, int r){
t[p].l = l, t[p].r = r;
if(l == r) {
t[p].sum = a[l];
return ;
}
int mid = (l + r) >> 1;
build(p*2, l, mid);
build(p*2+1, mid + 1, r);
t[p].sum = t[p*2].sum + t[p*2+1].sum;
}
void spread(int p){
if(t[p].add){
t[p*2].sum += (long long)(t[p*2].r - t[p*2].l + 1) * t[p].add;
t[p*2+1].sum += (long long)(t[p*2+1].r - t[p*2+1].l + 1) * t[p].add;
t[p*2].add += t[p].add;
t[p*2+1].add += t[p].add;
t[p].add = 0;
}
}
void change(int p, int l, int r, int val){
if(l <= t[p].l && r >= t[p].r){
t[p].add += val;
t[p].sum += (long long)(t[p].r - t[p].l + 1) * val;
return;
}
spread(p);
int mid = (t[p].l + t[p].r) >> 1;
if(l <= mid) change(p*2, l, r, val);
if(r > mid) change(p*2+1, l, r, val);
t[p].sum = t[p*2].sum + t[p*2+1].sum;
}
long long ask(int p, int l, int r){
if(l <= t[p].l && r >= t[p].r) return t[p].sum;
spread(p);
int mid = (t[p].l + t[p].r) >> 1;
long long val = 0;
if(l <= mid) val += ask(p*2, l, r);
if(r > mid) val += ask(p*2+1, l, r);
return val;
}
int main(){
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i++) cin >> a[i];
build(1, 1, n);
while(m--){
int f, x, y, h;
cin >> f >> x>> y;
if(f == 1){
cin >> h;
change(1, x, y, h);
}
if(f == 2){
cout << ask(1, x, y) <<"\n";
}
}
return 0;
}
并查集
板子不重要，路径压缩和带权才重要
int get(int x){
if(x == fa[x]) return x;
return fa[x] = get(fa[x]);
}
void merge(int x, int y){
fa[get(x)] = get(y);
}
平衡树Treap
#include<iostream>
#include<cstdlib>
using namespace std;
const int SIZE = 100005;
struct Treap{
int l, r;
int val, dat;
int size, cnt;
}a[SIZE];
int tot, root, n, INF=0x7fffffff;
int New(int val){
a[++tot].val = val;
a[tot].dat = rand();
a[tot].size = a[tot].cnt = 1;
return tot;
}
void Update(int p){
a[p].size = a[a[p].l].size + a[a[p].r].size + a[p].cnt;
}
void Build(){
New(-INF), New(INF);
root = 1, a[1].r = 2;
Update(root);
}
void zig(int &p){
int q = a[p].l;
a[p].l = a[q].r, a[q].r = p, p = q;
Update(a[p].r), Update(p);
}
void zag(int &p){
int q = a[p].r;
a[p].r = a[q].l, a[q].l = p, p = q;
Update(a[p].l), Update(p);
}
void Insert(int &p, int val){
if(p == 0){
p = New(val);
return;
}
if(a[p].val == val){
a[p].cnt++, Update(p);
return;
}
if(val < a[p].val){
Insert(a[p].l, val);
if(a[p].dat < a[a[p].l].dat) zig(p);
}
else{
Insert(a[p].r, val);
if(a[p].dat < a[a[p].r].dat) zag(p);
}
Update(p);
}
int GetRankByVal(int p, int val){
if(p == 0) return 0;
if(val == a[p].val) return a[a[p].l].size;
if(val < a[p].val) return GetRankByVal(a[p].l, val);
return a[a[p].l].size + a[p].cnt + GetRankByVal(a[p].r, val);
}
int GetValByRank(int p, int rank){
if(p == 0) return INF;
if(a[a[p].l].size >= rank) return GetValByRank(a[p].l, rank);
if(a[a[p].l].size + a[p].cnt >= rank) return a[p].val;
return GetValByRank(a[p].r, rank - a[a[p].l].size - a[p].cnt);
}
int GetPrev(int val){
int ans = 1;
int p = root;
while(p){
if(val == a[p].val){
if(a[p].l > 0){
p = a[p].l;
while(a[p].r > 0) p = a[p].r;
ans = p;
}
break;
}
if(a[ans].val < a[p].val && a[p].val < val) ans = p;
p = val < a[p].val ? a[p].l : a[p].r;
}
return a[ans].val;
}
int GetNext(int val){
int ans = 2;
int p = root;
while(p){
if(val == a[p].val){
if(a[p].r > 0){
p = a[p].r;
while(a[p].l > 0) p = a[p].l;
ans = p;
}
break;
}
if(a[ans].val > a[p].val && a[p].val > val) ans = p;
p = val < a[p].val ? a[p].l : a[p].r;
}
return a[ans].val;
}
void Remove(int &p, int val){
if(p == 0) return;
if(val == a[p].val){
if(a[p].cnt > 1){
a[p].cnt--, Update(p);
return;
}
if(a[p].l || a[p].r){
if(a[p].r == 0 && a[a[p].l].dat > a[a[p].r].dat)
zig(p), Remove(a[p].r, val);
else
zag(p), Remove(a[p].l, val);
Update(p);
}else p = 0;
return;
}
val < a[p].val ? Remove(a[p].l, val) : Remove(a[p].r, val);
Update(p);
}
int main(){
Build();
cin >> n;
while(n--){
int opt, x;
cin >> opt >> x;
switch(opt){
case 1:
Insert(root, x);
break;
case 2:
Remove(root, x);
break;
case 3:
cout << GetRankByVal(root, x) - 1 <<"\n";
break;
case 4:
cout << GetValByRank(root, x+1) <<"\n";
break;
case 5:
cout << GetPrev(x) <<"\n";
break;
case 6:
cout << GetNext(x) <<"\n";
break;
}
}
return 0;
}
平衡树Splay
#include<iostream>
using namespace std;
const int _ = 100005;
const int INF = 0x7fffffff;
int n, opt, x;
struct Spaly{
int root, tot, cnt[_], val[_], size[_], ch[_][2], fa[_];
bool chk(int x){
return ch[fa[x]][1] == x;
}
void pushup(int x){
if(!x) return;
size[x] = size[ch[x][0]] + size[ch[x][1]] + cnt[x];
}
int newnode(int x){
val[++tot] = x;
cnt[tot] = size[tot] = 1;
ch[x][0] = ch[x][1] = 0;
return tot;
}
void rotate(int x){
int y = fa[x], z = fa[y], k = chk(x), w = ch[x][k^1];
ch[y][k] = w, fa[w] = y;
ch[z][chk(y)] = x, fa[x] = z;
ch[x][k^1] = y, fa[y] = x;
pushup(y), pushup(x);
}
void splay(int x, int goal = 0){
while(fa[x] != goal){
int y = fa[x], z = fa[y];
if(z != goal){
if(chk(x) == chk(y)) rotate(y);
else rotate(x);
}
rotate(x);
}
if(!goal) root = x;
}
void insert(int x){
int cur = root, p = 0;
while(cur && x != val[cur])
p = cur, cur = ch[cur][x > val[cur]];
if(cur) ++cnt[cur], splay(cur);
else cur = ch[p][x > val[p]] = newnode(x), fa[cur] = p, splay(cur);
}
void find(int x){
if(!root) return;
int cur = root;
while(ch[cur][x > val[cur]] && x != val[cur]) cur = ch[cur][x > val[cur]];
splay(cur);
}
int rank(int x){
find(x);
return cnt[ch[root][0]] + 1;
}
int kth(int k){
int cur = root;
while(1){
if(ch[cur][0] && k <= size[ch[cur][0]]) cur = ch[cur][0];
else if(k > size[ch[cur][0]] + cnt[cur]) k -= size[ch[cur][0]] + cnt[cur], cur = ch[cur][1];
else return cur;
}
}
int pre(int x){
find(x);
if(val[root] < x) return root;
int cur = ch[root][0];
while(ch[cur][1]) cur = ch[cur][1];
return cur;
}
int next(int x){
find(x);
if(val[root] > x) return root;
int cur = ch[root][1];
while(ch[cur][0]) cur = ch[cur][0];
return cur;
}
void remove(int x){
int u = pre(x), v = next(x);
splay(u), splay(v, u);
int w = ch[v][0];
if(cnt[w] > 1) --cnt[w], splay(w);
else size[0] = cnt[0] = ch[v][0] = 0;
}
void maintain(int x){
if(!ch[x][0] && !ch[x][1]) size[x] = cnt[x];
if(ch[x][0]) maintain(ch[x][0]);
if(ch[x][1]) maintain(ch[x][1]);
pushup(x);
}
}T;
int main(){
T.insert(-INF), T.insert(INF);
cin >> n;
for(int i=1; i<=n; i++){
cin >> opt >> x;
if(opt == 1) T.insert(x);
if(opt == 2) T.remove(x);
if(opt == 3) cout << T.rank(x) - 1 <<"\n";
if(opt == 4) cout << T.val[T.kth(x)+1] <<"\n";
if(opt == 5) cout << T.val[T.pre(x)] <<"\n";
if(opt == 6) cout << T.val[T.next(x)] <<"\n";
}
return 0;
}
