算法模版

news/2025/10/14 23:31:26/文章来源:https://www.cnblogs.com/amlhdsan/p/19142328

\(算法模版\)

\(Author：\) 缪语博

本文档基于 \(GPL-3.0 License\)

本文档 \(GitHub\) 储存库：model

\(目录\) \(Contents\)

快读快写
线段树
中国剩余定理（拓展）
最短路算法
LCA最近公共祖先
高精度
树链剖分
网络流
KMP算法

基本格式

#include <bits/stdc++.h>using namespace std;int main() {freopen(".in", "r", stdin);freopen(".out", "w", stdout);return 0;
}

很好用的宏定义和函数

#define ll long long 
#define N 1000010
// 二分实现
#define MID ((l + r) >> 1)
// 线段树实现
#define PII pair<int, int>
#define MAX(x, y, z) max((x), max((y), (z)))
#define PRI priority_queue
//此处注意到底是几个零
#define MOD 1000000007
#define FI first
#define SE second
#define IL inline
#define RE register
#define MINN -0x7fffffff
#define MAXX 0x7fffffff
#define HMINN -0x3f3f3f3f
#define HMAXX 0x3f3f3f3f
#define ENDL putchar('\n')int nxt[N], head[N], to[N], to[N];
inline void add_edge(int u, int v, ll c) {nxt[++cnt] = head[u];head[u] = cnt;to[cnt] = v;w[cnt] = c;
}
// 遍历：
for(int i = head[p]; i; i = nxt[i]) {int v = to[i];// Write what you want to write.
}inline void madd(ll &a, ll b) {a += b;a %= MOD;
}

\(快读快写\)

inline int read() {int x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9') {if(ch == '-') {f = -1;}ch = getchar();}while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * f;
}
inline void write(int x) {if(x < 0) {putchar('-');x = -x;}if(x > 9)write(x / 10);putchar(x % 10 + '0');
}
inline string stringread() {string str = "";char ch = getchar();while(ch == ' ' || ch == '\n' || ch == '\r') {ch = getchar();}while(ch != ' ' && ch != '\n' && ch != '\r') {str += ch;ch = getchar();} return str;
}
inline void stringwrite(string str) {for(int i = 0; str[i] != '\0'; ++i)putchar(str[i]);
}

\(线段树\)

#include <bits/stdc++.h>#define N 100010
#define ll long long
#define ls (p << 1)
#define rs (p << 1 | 1)
#define mid ((l + r) >> 1)using namespace std;int n, m;
int a[N];
ll tree[N << 2];
int siz[N << 2];
int lazy[N << 2];void upd(int p) {tree[p] = tree[ls] + tree[rs];
}void upds(int p) {siz[p] = siz[ls] + siz[rs];
}void pushd(int p) {tree[ls] += lazy[p] * siz[ls];tree[rs] += lazy[p] * siz[rs];lazy[ls] += lazy[p];lazy[rs] += lazy[p];lazy[p] = 0;
}void build(int p, int l, int r) {lazy[p] = 0;if(l == r)  {siz[p] = 1;tree[p] = a[l];return ;}build(ls, l, mid);build(rs, mid + 1, r);upd(p);upds(p);
}void mdf(int p, int l, int r, int ql, int qr, int k) {if(ql <= l && r <= qr) {tree[p] += 1ll * siz[p] * k;lazy[p] += k;return;}pushd(p);if(ql <= mid) {mdf(ls, l, mid, ql, qr, k);}if(qr > mid) {mdf(rs, mid + 1, r, ql, qr, k);}upd(p);
}ll qry(int p, int l, int r, int ql, int qr) {if(ql <= l && r <= qr) {return tree[p];}pushd(p);ll sum = 0;if(ql <= mid) {sum += qry(ls, l, mid, ql, qr);}if(qr > mid) {sum += qry(rs, mid + 1, r, ql, qr);}return sum;
}int main() {cin >> n >> m;for(int i = 1; i <= n; ++i) {cin >> a[i];}build(1, 1, n);while(m--) {int op, x, y, k;cin >> op >> x >> y;if(op == 1) {cin >> k;mdf(1, 1, n, x, y, k);}else {cout << qry(1, 1, n, x, y) << endl;}}return 0;
}

\(中国剩余定理（拓展）\)

#include <bits/stdc++.h>#define N 100010
#define ll long longusing namespace std;ll n;
ll a[N], b[N];ll gcd(ll a, ll b) {if(!b) return a;else return gcd(b, a % b);
}inline ll read() {ll x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9') {if(ch == '-') {f = -1;}ch = getchar();}while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * f;
}void merge(ll p1, ll a1, ll p2, ll a2, ll &p, ll &a) {p = p1 / gcd(p1, p2) * p2;if(p1 < p2) {swap(p1, p2);swap(a1, a2);}a = a1;while(a % p2 != a2)a += p1;
}
int main() {n = read();for(int i = 1; i <= n; ++i) {a[i] = read();b[i] = read();b[i] %= a[i];}for(int i = 2; i <= n; ++i) {merge(a[i], b[i], a[i - 1], b[i - 1], a[i], b[i]);}cout << b[n] << endl;return 0;
}

\(最短路算法\)

\(Floyed\) 算法

#include <bits/stdc++.h>
using namespace std;int n, m;
int f[101][101];
int u, v, w;int main() {cin >> n >> m;for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j) {f[i][j] = 0x3f3f3f3f;if(i == j) {f[i][j] = 0;}}}for(int i = 1; i <= m; ++i) {cin >> u >> v >> w;f[u][v] = min(f[u][v], w);f[v][u] = f[u][v];}for(int k = 1; k <= n; ++k)for(int i = 1; i <= n; ++i)for(int j = 1; j <= n; ++j)f[i][j] = min(f[i][j], f[i][k] + f[k][j]);for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j) {cout << f[i][j] << ' ';}cout << endl;}return 0;
}

\(Dijkstra\) 算法

#include <bits/stdc++.h>
using namespace std;int n, m, s;
struct Edge {int v, c;
}tmp;
vector<Edge> edges[100005];
int dis[100005];
bool vis[100005];
priority_queue<pair<int, int> , vector<pair<int, int> >, greater<pair<int, int> > > pq;void dijkstra() {memset(dis, 0x3f, sizeof(dis));memset(vis, false, sizeof(vis));dis[s] = 0;pq.push(make_pair(0, s));while(!pq.empty()) {int u = pq.top().second;pq.pop();if(vis[u])continue;vis[u] = true;for(Edge e : edges[u]) {if(dis[e.v] > dis[u] + e.c) {dis[e.v] = dis[u] + e.c;pq.push(make_pair(dis[e.v], e.v));}}}
}int main() {cin >> n >> m >> s;for(int i = 1; i <= m; ++i) {int u, v, w;cin >> u >> v >> w;tmp.c = w;tmp.v = v;edges[u].push_back(tmp);}dijkstra();for(int i = 1; i <= n; ++i)cout << (dis[i] == 0x3f3f3f3f ? 2147483647 : dis[i]) << ' ';return 0;
}

\(Bellman-Ford\) 算法

#include <bits/stdc++.h>#define MAXN 100005using namespace std;int n, m, s;struct Edge {int v;int c;
};vector<Edge> edges[MAXN];
int dis[MAXN];inline int read() {int x = 0, f = 1;char ch = 0;while(ch < '0' || ch > '9') {if(ch == '-') {f = -1;}ch = getchar();}while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * f;
}inline void write(int x) {if(x < 0) {putchar('-');x = -x;}if(x > 9)write(x / 10);putchar(x % 10 + '0');
}void bellman_ford() {memset(dis, 0x3f, sizeof(dis));dis[s] = 0;bool flag;for(int i = 1; i <= n - 1; ++i) {flag = false;for(int u = 1; u <= n; ++u) {for(Edge e : edges[u]) {if(dis[e.v] > dis[u] + e.c) {dis[e.v] = dis[u] + e.c;flag = true;}}}if(!flag)break;}
}int main() {n = read();m = read();s = read();for(int i = 1; i <= m; ++i) {int u, v, c;u = read();v = read();c = read();Edge tmp;tmp.c = c;tmp.v = v;edges[u].push_back(tmp);}bellman_ford();for(int i = 1; i <= n; ++i) {write((dis[i] == 0x3f3f3f3f ? 2147483647 : dis[i]));putchar(' ');}putchar('\n');return 0;
}

\(LCA\) \(最近公共祖先\)

\(倍增写法：\)

#include <bits/stdc++.h>#define N 500005
#define ll long long
#define M 31using namespace std;int n, m, s;
int lg[N];
int xx, yy;
vector<int> edge[N];
int dep[N];
int fa[N][M];void add_edge(int u, int v) {edge[u].push_back(v);edge[v].push_back(u);
}void dfs(int p, int pre) {fa[p][0] = pre;dep[p] = dep[pre] + 1;for(int i = 1; i <= lg[dep[p]]; ++i) {fa[p][i] = fa[fa[p][i - 1]][i - 1];}for(int nxt : edge[p]) {if(nxt != pre) {dfs(nxt, p);}}
}int lca(int u, int v) {if(dep[u] < dep[v]) {swap(u, v);}while(dep[u] > dep[v]) {u = fa[u][lg[dep[u] - dep[v]] - 1];}if(u == v) {return u;}for(int llg = lg[dep[u]] - 1; llg >= 0; --llg) {if(fa[u][llg] != fa[v][llg]) {u = fa[u][llg];v = fa[v][llg];}}return fa[u][0];
}int main() {cin >> n >> m >> s;for(int i = 1; i <= n - 1; ++i) {cin >> xx >> yy;add_edge(xx, yy);}int cntt = 0;for(int i = 1; i <= n; ++i) {if(i >= (1 << cntt)) {++cntt;lg[i] = cntt;}else {lg[i] = cntt;}}dfs(s, 0);while(m--) {cin >> xx >> yy;cout << lca(xx, yy) << endl;}return 0;
}

\(高精度\)

int compare(string str1,string str2)
{if(str1.length()>str2.length()) return 1;else if(str1.length()<str2.length())  return -1;else return str1.compare(str2);
}//高精度加法
//只能是两个正数相加
string add(string str1,string str2)
{string str;int len1=str1.length();int len2=str2.length();//前面补0，弄成长度相同if(len1<len2){for(int i=1;i<=len2-len1;i++)str1="0"+str1;}else{for(int i=1;i<=len1-len2;i++)str2="0"+str2;}len1=str1.length();int cf=0;int temp;for(int i=len1-1;i>=0;i--){temp=str1[i]-'0'+str2[i]-'0'+cf;cf=temp/10;temp%=10;str=char(temp+'0')+str;}if(cf!=0)  str=char(cf+'0')+str;return str;
}//高精度减法
//只能是两个正数相减，而且要大减小
string sub(string str1,string str2)
{string str;int tmp=str1.length()-str2.length();int cf=0;for(int i=str2.length()-1;i>=0;i--){if(str1[tmp+i]<str2[i]+cf){str=char(str1[tmp+i]-str2[i]-cf+'0'+10)+str;cf=1;}else{str=char(str1[tmp+i]-str2[i]-cf+'0')+str;cf=0;}}for(int i=tmp-1;i>=0;i--){if(str1[i]-cf>='0'){str=char(str1[i]-cf)+str;cf=0;}else{str=char(str1[i]-cf+10)+str;cf=1;}}str.erase(0,str.find_first_not_of('0'));//去除结果中多余的前导0return str;
}//高精度乘法
//只能是两个正数相乘
string mul(string str1,string str2)
{string str;int len1=str1.length();int len2=str2.length();string tempstr;for(int i=len2-1;i>=0;i--){tempstr="";int temp=str2[i]-'0';int t=0;int cf=0;if(temp!=0){for(int j=1;j<=len2-1-i;j++)tempstr+="0";for(int j=len1-1;j>=0;j--){t=(temp*(str1[j]-'0')+cf)%10;cf=(temp*(str1[j]-'0')+cf)/10;tempstr=char(t+'0')+tempstr;}if(cf!=0) tempstr=char(cf+'0')+tempstr;}str=add(str,tempstr);}str.erase(0,str.find_first_not_of('0'));return str;
}//高精度除法
//两个正数相除，商为quotient,余数为residue
//需要高精度减法和乘法
void div(string str1,string str2,string &quotient,string &residue)
{quotient=residue="";//清空if(str2=="0")//判断除数是否为0{quotient=residue="ERROR";return;}if(str1=="0")//判断被除数是否为0{quotient=residue="0";return;}int res=compare(str1,str2);if(res<0){quotient="0";residue=str1;return;}else if(res==0){quotient="1";residue="0";return;}else{int len1=str1.length();int len2=str2.length();string tempstr;tempstr.append(str1,0,len2-1);for(int i=len2-1;i<len1;i++){tempstr=tempstr+str1[i];tempstr.erase(0,tempstr.find_first_not_of('0'));if(tempstr.empty())tempstr="0";for(char ch='9';ch>='0';ch--)//试商{string str,tmp;str=str+ch;tmp=mul(str2,str);if(compare(tmp,tempstr)<=0)//试商成功{quotient=quotient+ch;tempstr=sub(tempstr,tmp);break;}}}residue=tempstr;}quotient.erase(0,quotient.find_first_not_of('0'));if(quotient.empty()) quotient="0";
}

\(树链剖分\)

#include <bits/stdc++.h>#define N 200010
#define ENDL putchar('\n')#define mid ((l + r) >> 1)
#define ls (p << 1)
#define rs (p << 1 | 1)using namespace std;int n, m, r, MOD;
int nxt[N], head[N], to[N], w[N];
int top[N];
int si[N], fa[N], id[N], son[N], wt[N], dep[N];
int cnt = 0, e = 0;
int tree[N << 2], siz[N << 2], lazy[N << 2];inline int read() {int x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9') {if(ch == '-') {f = -1;}ch = getchar();}while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * f;
}inline void write(int x) {if(x < 0) {putchar('-');x = -x;}if(x > 9)write(x / 10);putchar(x % 10 + '0');
}inline void add(int &a, int b) {a = (a + b) % MOD;
}inline void upd(int p) {tree[p] = (tree[ls] + tree[rs]) % MOD;
}inline void upds(int p) {siz[p] = siz[ls] + siz[rs];
}inline void pushd(int p) {if(!lazy[p])return;add(tree[ls], siz[ls] * lazy[p] % MOD);add(tree[rs], siz[rs] * lazy[p] % MOD);add(lazy[ls], lazy[p]);add(lazy[rs], lazy[p]);lazy[p] = 0;
}inline void build(int p, int l, int r) {if(l == r) {tree[p] = wt[l];siz[p] = 1;lazy[p] = 0;return;}build(ls, l, mid);build(rs, mid + 1, r);upd(p);upds(p);
}inline void mdf(int p, int l, int r, int ql, int qr, int x) {if(ql <= l && r <= qr) {add(tree[p], siz[p] * x % MOD);add(lazy[p], x);return;}pushd(p);if(ql <= mid) {mdf(ls, l, mid, ql, qr, x);}if(qr > mid) {mdf(rs, mid + 1, r, ql, qr, x);}upd(p);
}inline int qry(int p, int l, int r, int ql, int qr) {if(ql <= l && r <= qr) {return tree[p];}pushd(p);int sum = 0;if(ql <= mid) {add(sum, qry(ls, l, mid, ql, qr));}if(qr > mid) {add(sum, qry(rs, mid + 1, r, ql, qr));}return sum;
}inline void add_edge(int u, int v) {nxt[++e] = head[u];head[u] = e;to[e] = v;
}inline void dfs1(int p, int pre, int depth) {dep[p] = depth;fa[p] = pre;si[p] = 1;int maxx = -1;for(int i = head[p]; i; i = nxt[i]) {int v = to[i];if(v != pre) {dfs1(v, p, depth + 1);si[p] +=  si[v];if(si[v] > maxx) {maxx = si[v];son[p] = v;}}}
}inline void dfs2(int p, int topp) {id[p] = ++cnt;wt[cnt] = w[p];top[p] = topp;if(!son[p])return;dfs2(son[p], topp);for(int i = head[p]; i; i = nxt[i]) {int v = to[i];if(v != son[p] && v != fa[p]) {dfs2(v, v);}}
}inline void modify1(int l, int r, int x) {x %= MOD;while(top[l] != top[r]) {if(dep[top[l]] < dep[top[r]]) {swap(l, r);}mdf(1, 1, n, id[top[l]], id[l], x);l = fa[top[l]];}if(dep[l] > dep[r])swap(l, r);mdf(1, 1, n, id[l], id[r], x);
}inline int query1(int l, int r) {int sum = 0;while(top[l] != top[r]) {if(dep[top[l]] < dep[top[r]]) {swap(l, r);}add(sum, qry(1, 1, n, id[top[l]], id[l]));l = fa[top[l]];}if(dep[l] > dep[r])swap(l, r);add(sum, qry(1, 1, n, id[l], id[r]));return sum;
}inline void modify2(int p, int x) {mdf(1, 1, n, id[p], id[p] + si[p] - 1, x);
}inline int query2(int p) {return qry(1, 1, n, id[p], id[p] + si[p] - 1) % MOD;
}int main() {n = read();m = read();r = read();MOD = read();for(int i = 1; i <= n; ++i) {w[i] = read();}for(int i = 1; i <= n - 1; ++i) {int u, v;u = read();v = read();add_edge(u, v);add_edge(v, u);}dfs1(r, 0, 1);dfs2(r, r);build(1, 1, n);while(m--) {int op;op = read();if(op == 1) {int x, y, z;x = read();y = read();z = read();modify1(x, y, z);}else if(op == 2) {int x, y;x = read();y = read();write(query1(x, y));ENDL;}else if(op == 3) {int x, z;x = read();z = read();modify2(x, z);}else {int x;x = read();write(query2(x));ENDL;}}return 0;
}

\(网络流\)

\(Dinic\) 最大流

#include <bits/stdc++.h>#define N 500010using namespace std;long long nxt[N], to[N], head[N], w[N];
long long now[N];
int d[N];
long long n, m, s, t;
long long cnt = 1;inline void add(int u, int v, long long c) {nxt[++cnt] = head[u];head[u] = cnt;to[cnt] = v;w[cnt] = c;nxt[++cnt] = head[v];head[v] = cnt;to[cnt] = u;w[cnt] = 0;
}inline bool bfs() {queue<int> q;q.push(s);memset(d, 0x3f, sizeof(d));d[s] = 0;now[s] = head[s];while(!q.empty()) {int x = q.front();q.pop();for(int i = head[x]; i; i = nxt[i]) {int v = to[i];if(d[v] == 0x3f3f3f3f && w[i] > 0) {d[v] = d[x] + 1;now[v] = head[v];q.push(v);if(v == t) {return 1;}}}}return 0;
}inline long long dfs(int x, long long flow) {if(x == t) {return flow;}long long k;for(int i = now[x]; i; i = nxt[i]) {now[x] = i;int v = to[i];if(d[v] != d[x] + 1 || w[i] <= 0) {continue;}k = dfs(v, min(flow, w[i]));if(k) {w[i] -= k;w[i ^ 1] += k;return k;}else {d[v] = 0x3f3f3f3f;}}return 0;
}inline long long dinic() {long long sum = 0;while(bfs()) {sum += dfs(s, 0x7fffffff);}return sum;
}int main() {cin >> n >> m >> s >> t;while(m--) {long long u, v, c;cin >> u >> v >> c;add(u, v, c);}cout << dinic() << endl;return 0;
}

\(KMP\) \(算法\)

#include <bits/stdc++.h>#define N 1000001using namespace std;char s1[N], s2[N];
int la, lb;
int j = 0;
int kmp[N];int main() {cin >> s1 + 1;cin >> s2 + 1;la = strlen(s1 + 1);lb = strlen(s2 + 1);for(int i = 1; i <= lb; ++i) {while(j && s2[i + 1] != s2[j + 1]) {j = kmp[j];}if(s2[i + 1] == s2[j + 1]) {++j;kmp[i + 1] = j;}}j = 0;for(int i = 1; i <= la; ++i) {while(j && s1[i] != s2[j + 1]) {j = kmp[j];}if(s1[i] == s2[j + 1]) {++j;}if(j == lb) {cout << i - lb + 1 << endl;j = kmp[j];}}for(int i = 1; i <= lb; ++i) {cout << kmp[i] << ' ';}return 0;
}

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newDay10

1.今天事有点多，课后没学多少，主要是写了写作业，当然单词还是得背的。明天应该有空了 2.明天多花点时间学学 3.太粗心了，有点改不过来