模板整理 (施工中 2017.8.21更新)
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目录
数据结构
1 字典树
图论
1 网络流dinic
2 zkw费用流
3 无向图的强联通分量
数论
1 中国剩余定理
2 大质数判定
3 约瑟夫环
其他
1 祖传头文件
2 大数
数据结构
1.字典树
使用时记得先初始化第一个点
const int MAXN = 1e2 + 5;
int Maxlen, Maxnum;
int T, K;
struct node {
int next[27];
int v, num;
void init() {
v=-1;
num = 0;
memset(next,-1,sizeof(next));
}
};
struct node L[1000000];
int tot=0;
void add(char a[]) {
int now=0;
int len = strlen(a);
for(int i=0; i<len; i++) {
int tmp=a[i]-‘a‘;
int next=L[now].next[tmp];
if(next==-1) {
next=++tot;
L[next].init();
L[now].next[tmp]=next;
}
now=next;
L[now].num ++;
if(L[now].num >= 3) {
if(Maxlen < i + 1) {
Maxlen = i + 1;
Maxnum = L[now].num;
} else if(Maxnum < L[now].num && Maxlen <= i + 1) {
Maxnum = L[now].num;
}
}
}
L[now].v=0;
}
int query(char s2[]) {
int len = strlen(s2);
int now = 0;
for(int i = 0; i < len; i++) {
int temp = s2[i] - ‘a‘;
int next = L[now].next[temp];
if(next == -1) return 0;
now = next;
}
return L[now].num;
}
int main()
{
L[0].init();
return 0;
}
图论
1.网络流Dinic
const int maxn = 405; //开四倍
const int maxe = 4*maxn*maxn;
const int inf = 0x3f3f3f3f;
struct MaxFlow {
struct Edge {
int v, w, nxt;
} edge[maxe];
int head[maxn], tot, level[maxn];
void init(){
memset(head,-1,sizeof(head));
tot=0;
}
void add(int u, int v, int w) {
edge[tot].v = v;
edge[tot].w = w;
edge[tot].nxt = head[u];
head[u] = tot++;
edge[tot].v = u;
edge[tot].w = 0;
edge[tot].nxt = head[v];
head[v] = tot++;
}
bool bfs(int s, int t) {
memset(level, -1, sizeof(level));
queue<int>q;
q.push(s);
level[s] = 0;
while(!q.empty()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = edge[i].nxt) {
if(edge[i].w > 0 && level[edge[i].v] < 0) {
level[edge[i].v] = level[u] + 1;
q.push(edge[i].v);
}
}
}
return level[t] > 0;
}
int dfs(int u, int t, int f) {
if(u == t) return f;
for(int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if(edge[i].w > 0 && level[v] > level[u]) {
int d = dfs(v, t, min(f, edge[i].w));
if(d > 0) {
edge[i].w -= d;
edge[i ^ 1].w += d;
return d;
}
}
}
level[u] = -1; //不太确定,如果WA了把这句删掉试试
return 0;
}
int solve(int s, int t) {
int flow = 0, f;
while(bfs(s, t)) {
while(f = dfs(s, t, inf)) flow += f;
}
return flow;
}
}F;
2.
zkw费用流
const int MX = 80000;
const int MXE = 4 * MX * MX;
const int ME = 4e5 + 5;//边的数量
struct MCMF {
int S, T;//源点，汇点
int tot, n;
int st, en, maxflow, mincost;
bool vis[MX];
int head[MX], cur[MX], dis[MX];
int roade[MX], roadv[MX], rsz; //用于打印路径
queue <int> Q;
struct Edge {
int v, cap, cost, nxt, flow;
Edge() {}
Edge(int a, int b, int c, int d) {
v = a, cap = b, cost = c, nxt = d, flow = 0;
}
} E[ME], SE[ME];
void init(int _n) {
n = _n, tot = 0;
for(int i = 0; i <= n; i++) head[i] = -1;
}
void edge_add(int u, int v, int cap, int cost) {
E[tot] = Edge(v, cap, cost, head[u]);
head[u] = tot++;
E[tot] = Edge(u, 0, -cost, head[v]);
head[v] = tot++;
}
bool adjust() {
int v, min = INF;
for(int i = 0; i <= n; i++) {
if(!vis[i]) continue;
for(int j = head[i]; ~j; j = E[j].nxt) {
v = E[j].v;
if(E[j].cap - E[j].flow) {
if(!vis[v] && dis[v] - dis[i] + E[j].cost < min) {
min = dis[v] - dis[i] + E[j].cost;
}
}
}
}
if(min == INF) return false;
for(int i = 0; i <= n; i++) {
if(vis[i]) {
cur[i] = head[i];
vis[i] = false;
dis[i] += min;
}
}
return true;
}
int augment(int i, int flow) {
if(i == en) {
mincost += dis[st] * flow;
maxflow += flow;
return flow;
}
vis[i] = true;
for(int j = cur[i]; j != -1; j = E[j].nxt) {
int v = E[j].v;
if(E[j].cap == E[j].flow) continue;
if(vis[v] || dis[v] + E[j].cost != dis[i]) continue;
int delta = augment(v, std::min(flow, E[j].cap - E[j].flow));
if(delta) {
E[j].flow += delta;
E[j ^ 1].flow -= delta;
cur[i] = j;
return delta;
}
}
return 0;
}
void spfa() {
int u, v;
for(int i = 0; i <= n; i++) {
vis[i] = false;
dis[i] = INF;
}
Q.push(st);
dis[st] = 0; vis[st] = true;
while(!Q.empty()) {
u = Q.front(), Q.pop(); vis[u] = false;
for(int i = head[u]; ~i; i = E[i].nxt) {
v = E[i].v;
if(E[i].cap == E[i].flow || dis[v] <= dis[u] + E[i].cost) continue;
dis[v] = dis[u] + E[i].cost;
if(!vis[v]) {
vis[v] = true;
Q.push(v);
}
}
}
for(int i = 0; i <= n; i++) {
dis[i] = dis[en] - dis[i];
}
}
int zkw(int s, int t) {
st = s, en = t;
spfa();
mincost = maxflow = 0;
for(int i = 0; i <= n; i++) {
vis[i] = false;
cur[i] = head[i];
}
do {
while(augment(st, INF)) {
memset(vis, false, n * sizeof(bool));
}
} while(adjust());
return mincost;
}
} M;
3.无向图的强联通分量
const int MX = 1e5 + 10;
struct Edge {
int u, v, nxt;
} E[MX];
int Head[MX], erear;
void edge_init() {
erear = 0;
memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v) {
E[erear].u = u;
E[erear].v = v;
E[erear].nxt = Head[u];
Head[u] = erear++;
}
int n, m, IN[MX], cnt[MX], val[MX];
int bsz, ssz, dsz;
int Low[MX], DFN[MX];
void Init_tarjan(int n) {
bsz = ssz = dsz = 0;
for(int i = 1; i <= n; ++i) Low[i] = DFN[i] = 0;
}
int Stack[MX], inStack[MX], Belong[MX];
void trajan(int u, int e) {
inStack[u] = 1;
Stack[++ssz] = u;
DFN[u] = Low[u] = ++dsz;
for(int i = Head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
if((i ^ 1) == e) continue;
if(!DFN[v]) {
trajan(v, i);
Low[u] = min(Low[u], Low[v]);
} else if(inStack[v]) {
Low[u] = min(Low[u], Low[v]);
}
}
if(DFN[u] == Low[u]) {
bsz++; int v;
do {
v = Stack[ssz--];
inStack[v] = 0;
Belong[v] = bsz;
} while(ssz && v != u);
}
}
void tarjan_solve(int n) {
dsz = bsz = ssz = 0;
memset(DFN, 0, sizeof(DFN));
for(int i = 1; i <= n; i++) {
if(!DFN[i]) trajan(i, -1);
}
/*缩点*/
edge_init();
for(int i = 0; i < 2 * m; i += 2) {
int u = E[i].u, v = E[i].v;
u = Belong[u]; v = Belong[v];
if(u == v) continue;
edge_add(u, v);
edge_add(v, u);
}
}
数论
1.中国剩余定理
LL gcd(LL a, LL b, LL &x, LL &y) {
if(b == 0) {
x =1, y = 0;
return a;
}
LL r = gcd(b, a % b, x, y);
LL t = y;
y = x - a / b * y;
x = t;
return r;
}
LL multi(LL a, LL b, LL mod) {
LL ret = 0;
while(b) {
if(b & 1) {
ret = ret + a;
if(ret >= mod) ret -= mod;
}
a = a + a;
if(a >= mod) a -= mod;
b >>= 1;
}
return ret;
}
LL crt(int n, LL m[], LL a[]) {
LL M = 1, d, y, x = 0;
for(int i = 0; i < n; i++) M *= m[i];
for(int i = 0; i < n; i++) {
LL w = M / m[i];
d = gcd(m[i], w, d, y);
y = (y % m[i] + m[i]) % m[i];
x = ((x + multi(multi(a[i], w, M), y, M)) % M + M) % M;
}
return x;
}
2.大质数判定
LL multi(LL a, LL b, LL mod) {
LL ret = 0;
while(b) {
if(b & 1) ret = ret + a;
if(ret >= mod) ret -= mod;
a = a + a;
if(a >= mod) a -= mod;
b >>= 1;
}
return ret;
}
LL power(LL a, LL b, LL mod) {
LL ret = 1;
while(b) {
if(b & 1) ret = multi(ret, a, mod);
a = multi(a, a, mod);
b >>= 1;
}
return ret;
}
bool Miller_Rabin(LL n) {
LL u = n - 1, pre, x;
int i, j, k = 0;
if(n == 2 || n == 3 || n == 5 || n == 7 || n == 11) return true;
if(n == 1 || (!(n % 2)) || (!(n % 3)) || (!(n % 5)) || (!(n % 7)) || (!(n % 11))) return
false;
for(; !(u & 1); k++, u >>= 1);
for(i = 0; i < 5; i++) {
x = rand() % (n - 2) + 2;
x = power(x, u, n);
pre = x;
for(j = 0; j < k; j++) {
x = multi(x, x, n);
if(x == 1 && pre != 1 && pre != (n - 1))
return false;
pre = x;
}
if(x != 1) return false;
}
return true;
}
3.约瑟夫环
/*
F[n] = (F[n - 1] + m) % n, F[1] = 0
返回的下标从0 开始，复杂度大约为O(m)
*/
int Joseph(int n, int m) {
if(n == 1) return 0;
if(m == 1) return n - 1;
LL pre = 0; int now = 2;
while(now <= n) {
if(pre + m >= now) {
pre = (pre + m) % now;
now++;
} else {
int a = now - 1 - pre, b = m - 1;
int k = a / b + (a % b != 0);
if(now + k > n + 1) k = n + 1 - now;
pre = (pre + (LL)m * k) % (now + k - 1);
now += k;
}
}
return pre;
}
其他
1.祖传头文件
#include <iostream> #include <cstring> #include <stdio.h> #include <algorithm> #include <queue> #include <vector> #include <iomanip> #include <math.h> #include <map> using namespace std; #define FIN freopen("input.txt","r",stdin); #define FOUT freopen("output.txt","w",stdout); #define INF 0x3f3f3f3f #define INFLL 0x3f3f3f3f3f3f3f #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef long long LL; typedef pair<int, int> PII;
2.大数
const int MX = 2500;
const int MAXN = 9999;
const int DLEN = 4;
/*已重载>+- % 和print*/
class Big {
public:
int a[MX], len;
Big(const int b = 0) {
int c, d = b;
len = 0;
memset(a, 0, sizeof(a));
while(d > MAXN) {
c = d - (d / (MAXN + 1)) * (MAXN + 1);
d = d / (MAXN + 1);
a[len++] = c;
}
a[len++] = d;
}
Big(const char *s) {
int t, k, index, L, i;
memset(a, 0, sizeof(a));
L = strlen(s);
len = L / DLEN;
if(L % DLEN) len++;
index = 0;
for(i = L - 1; i >= 0; i -= DLEN) {
t = 0;
k = i - DLEN + 1;
if(k < 0) k = 0;
for(int j = k; j <= i; j++) {
t = t * 10 + s[j] - ‘0‘;
}
a[index++] = t;
}
}
Big operator/(const int &b)const {
Big ret;
int i, down = 0;
for(int i = len - 1; i >= 0; i--) {
ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
return ret;
}
bool operator>(const Big &T)const {
int ln;
if(len > T.len) return true;
else if(len == T.len) {
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0) ln--;
if(ln >= 0 && a[ln] > T.a[ln]) return true;
else return false;
} else return false;
}
Big operator+(const Big &T)const {
Big t(*this);
int i, big;
big = T.len > len ? T.len : len;
for(i = 0; i < big; i++) {
t.a[i] += T.a[i];
if(t.a[i] > MAXN) {
t.a[i + 1]++;
t.a[i] -= MAXN + 1;
}
}
if(t.a[big] != 0) t.len = big + 1;
else t.len = big;
return t;
}
Big operator-(const Big &T)const {
int i, j, big;
bool flag;
Big t1, t2;
if(*this > T) {
t1 = *this;
t2 = T;
flag = 0;
} else {
t1 = T;
t2 = *this;
flag = 1;
}
big = t1.len;
for(i = 0; i < big; i++) {
if(t1.a[i] < t2.a[i]) {
j = i + 1;
while(t1.a[j] == 0) j++;
t1.a[j--]--;
while(j > i) t1.a[j--] += MAXN;
t1.a[i] += MAXN + 1 - t2.a[i];
} else t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[t1.len - 1] == 0 && t1.len > 1) {
t1.len--;
big--;
}
if(flag) t1.a[big - 1] = 0 - t1.a[big - 1];
return t1;
}
int operator%(const int &b)const {
int i, d = 0;
for(int i = len - 1; i >= 0; i--) {
d = ((d * (MAXN + 1)) % b + a[i]) % b;
}
return d;
}
Big operator*(const Big &T) const {
Big ret;
int i, j, up, temp, temp1;
for(i = 0; i < len; i++) {
up = 0;
for(j = 0; j < T.len; j++) {
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > MAXN) {
temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = temp1;
} else {
up = 0;
ret.a[i + j] = temp;
}
}
if(up != 0) {
ret.a[i + j] = up;
}
}
ret.len = i + j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
return ret;
}
void print() {
printf("%d", a[len - 1]);
for(int i = len - 2; i >= 0; i--) printf("%04d", a[i]);
}
};
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