$\rm{NOIp}$板子整理

其他 2018-11-10 09:24:34 阅读次数: 0

一、并查集的两种方式

其实就是一个随机化路径压缩，一个启发式合并。

#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <iostream>

#define MAXN 200010

using namespace std ;
int N, M, A, B, C, F[MAXN] ;

namespace _Heuristic{
    int H[MAXN] ;
    int find(int x){
        if (x == F[x]) return x ;
        return F[x] = find(F[x]) ;
    }
    inline void Merge(int x, int y){ // Heuristic Merge 
        int f1 = find(x), f2 = find(y) ; // Insert the less-height tree into the higher tree
        if (H[f1] > H[f2]) F[f2] = f1 ; //because : H[x] < H[y] -> After Insert H[x] to H[y], H[y] ++ or H[y] == ;
        else if (H[f1] < H[f2]) F[f1] = f2 ; //Therefore, we make the tree the lowest height under meaning .
        else F[f1] = f2, ++ H[f2] ;
    }
    void solve2(){
        fill(H, H + N + 2, 1) ;
        while (M --){
            scanf("%d%d%d", &A, &B, &C) ;
            if (A == 2){
                int f1 = find(B), f2 = find(C) ;
                printf("%c\n", f1 == f2 ? 'Y' : 'N') ;
            }
            else Merge(B, C) ; 
        }
    }
}
namespace _Compress{
    int t ;
    int find(int x){
        if (x == F[x]) return x ;
        return F[x] = find(F[x]) ;
    }
    void solve1(){
        srand(time(NULL)) ; 
        while (M --){
            scanf("%d%d%d", &A, &B, &C) ;
            if (A == 2){
                int f1 = find(B), f2 = find(C) ;
                printf("%c\n", f1 == f2 ? 'Y' : 'N') ;
            }
            else {
                int f1 = find(B), f2 = find(C) ;
                if (!(t = rand() % 2)) F[f1] = f2 ; else F[f2] = f1 ;
            }   
        }
    }
}
int main(){
    cin >> N >> M ;  
    for (int i = 1 ; i <= N ; ++ i) F[i] = i  ;
    if (M <= 50000) _Compress::solve1() ; else _Heuristic::solve2() ; return 0 ;
}

二、筛素数的两种方式

其实就是埃氏筛和线性筛啦~

#include <bitset>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std ;
int N, M ; 

namespace Es{ // Eratosthenes
    #define MAXN 2000100
    bool vis[MAXN] ; int A, i, j ;
    void Ego(){
        vis[1] = vis[0] = 1 ;
        for (i = 2 ; i <= N ; ++ i){
            if (vis[i])  continue ; 
            for (j = i + i ; j <= N ; j += i) vis[j] = 1 ;
        }
    }
    void solve1(){
        Ego() ;
        while (M --)
            scanf("%d", &A), printf("%s\n", !vis[A] ? "Yes" : "No") ;
    }
    #undef MAXN 
}
namespace Euler{
    #define MAXN 10001000
    bitset <MAXN> vis ;
    int Prime[MAXN], A, i, j, cnt ;
    void Ego(){
        vis[1] = vis[0] = 1 ;
        for (i = 2 ; i <= N ; ++ i){
            if (!vis[i]) Prime[++ cnt] = i ;
            for (j = 1 ; j <= cnt ; ++ j){
                if (i * Prime[j] > N) break ;
                vis[i * Prime[j]] = 1 ;
                if (!(i % Prime[j])) break ;
            }
        }
    }
    void solve2(){
        Ego() ;
        while (M --)
            scanf("%d", &A), printf("%s\n", !vis[A] ? "Yes" : "No") ;
    }
}
int main(){
    cin >> N >> M ;
    if (N <= 1000000) Es::solve1() ; else Euler::solve2() ; return 0 ; 
}

三、最短路的两种方式

\(SPFA + SLF\)+乱搞

#include <deque>
#include <cstdio>
#include <iostream>
 
#define R register 
#define MAXN 200010
#define to(i) E[i].to
#define Inf 2147483647

using namespace std ;
deque <int> q ;
struct Edge{
    int to, next, v ;
}E[MAXN] ; int head[MAXN], cnt ;
int N, M, S, dist[MAXN] ; bool vis[MAXN] ;

inline int qr(){
    int res = 0 ; char c = getchar() ;
    while (!isdigit(c)) c = getchar() ;
    while (isdigit(c)) res = (res << 1) + (res << 3) + c - 48, c = getchar() ;
    return res ; 
}
inline void Add(int u, int v, int w) {
    E[++ cnt].to = v, E[cnt].v = w ;
    E[cnt].next = head[u], head[u] = cnt ;
}
inline void SPFA(){
    for (R int i = 1 ; i <= N ; ++ i) dist[i] = Inf ; 
    dist[S] = 0, q.push_back(S), vis[S] = 1 ;
    while(!q.empty()){
        R int now = q.front() ; 
        q.pop_front(), vis[now] = 0 ;
        for (R int i = head[now] ; i ; i = E[i].next){
            if (dist[to(i)] > dist[now] + E[i].v){
                dist[to(i)] = dist[now] + E[i].v ;
                if (!vis[to(i)]){
                    vis[to(i)] = 1 ;
                    R int F = q.front(), B = q.back() ;
                    if (dist[F] < dist[B]){
                        q.pop_front(), q.pop_back() ;
                        q.push_front(B), q.push_back(F) ;
                        if (dist[B] > dist[to(i)])
                            q.push_front(to(i)) ;
                        else q.push_back(to(i)) ;
                    }
                    else if (dist[F] > dist[to(i)])
                            q.push_front(to(i)) ;
                        else q.push_back(to(i)) ; 
                }
            }
        }
    }
} 
int main(){
    cin >> N >> M >> S ; int A, B, C ;
    for (R int i = 1 ; i <= M ; ++ i) A = qr(), B = qr(), C = qr(), Add(A, B, C) ;
    SPFA() ; for (R int i = 1 ; i <= N ; ++ i) printf("%d ", dist[i]) ; return 0 ;
}

\(Heap + dijkstra\)

#include <deque>
#include <cstdio>
#include <iostream>
 
#define R register 
#define MAXN 200010
#define to(i) E[i].to
#define Inf 2147483647

using namespace std ;
deque <int> q ;
struct Edge{
    int to, next, v ;
}E[MAXN] ; int head[MAXN], cnt ;
int N, M, S, dist[MAXN] ; bool vis[MAXN] ;

inline int qr(){
    int res = 0 ; char c = getchar() ;
    while (!isdigit(c)) c = getchar() ;
    while (isdigit(c)) res = (res << 1) + (res << 3) + c - 48, c = getchar() ;
    return res ; 
}
inline void Add(int u, int v, int w) {
    E[++ cnt].to = v, E[cnt].v = w ;
    E[cnt].next = head[u], head[u] = cnt ;
}
inline void SPFA(){
    for (R int i = 1 ; i <= N ; ++ i) dist[i] = Inf ; 
    dist[S] = 0, q.push_back(S), vis[S] = 1 ;
    while(!q.empty()){
        R int now = q.front() ; 
        q.pop_front(), vis[now] = 0 ;
        for (R int i = head[now] ; i ; i = E[i].next){
            if (dist[to(i)] > dist[now] + E[i].v){
                dist[to(i)] = dist[now] + E[i].v ;
                if (!vis[to(i)]){
                    vis[to(i)] = 1 ;
                    R int F = q.front(), B = q.back() ;
                    if (dist[F] < dist[B]){
                        q.pop_front(), q.pop_back() ;
                        q.push_front(B), q.push_back(F) ;
                        if (dist[B] > dist[to(i)])
                            q.push_front(to(i)) ;
                        else q.push_back(to(i)) ;
                    }
                    else if (dist[F] > dist[to(i)])
                            q.push_front(to(i)) ;
                        else q.push_back(to(i)) ; 
                }
            }
        }
    }
} 
int main(){
    cin >> N >> M >> S ; int A, B, C ;
    for (R int i = 1 ; i <= M ; ++ i) A = qr(), B = qr(), C = qr(), Add(A, B, C) ;
    SPFA() ; for (R int i = 1 ; i <= N ; ++ i) printf("%d ", dist[i]) ; return 0 ;
}

四、矩阵快速幂

主要是成员函数的写法

#include <cstdio>
#include <cstring>
#include <iostream>

#define LL long long
#define Mod 1000000007

using namespace std ;
LL N, K ;
struct Matrix{
    LL M[200][200] ;
    void clear() { memset(M, 0, sizeof(M)) ;}
    void reset() {
        clear() ;
        for (int i = 1 ; i <= N ; ++ i) M[i][i] = 1 ;
    }
    Matrix friend operator *(const Matrix&A, const Matrix &B){
        Matrix Ans ; Ans.clear() ; 
        for (int i = 1 ; i <= N; ++ i)
            for (int j = 1 ; j <= N ; ++ j)
                for (int k = 1 ; k <= N; ++ k)
                    Ans.M[i][j] = (Ans.M[i][j] + A.M[i][k] * B.M[k][j]) % Mod ;
        return Ans ;
    }
    Matrix friend operator +(const Matrix&A, const Matrix &B){
        Matrix Ans ; Ans.clear() ; 
        for (int i = 1 ; i <= N; ++ i)
            for (int j = 1 ; j <= N ; ++ j)
                    Ans.M[i][j] = (A.M[i][j] + B.M[i][j]) % Mod ;
        return Ans ;
    }
} qwq, unit ; 
inline Matrix expow(Matrix T, LL P){
    Matrix Ans ; Ans.reset() ;
    while (P){
        if (P & 1) Ans = Ans * T ;
        T = T * T, P >>= 1 ;
    }
    return Ans ;
}
int main(){
    cin >> N >> K ;
    for (int i = 1 ; i <= N ; ++ i)
        for (int j = 1 ; j <= N ; ++ j)
            cin >> qwq.M[i][j] ;
    qwq = expow(qwq, K) ;
    for (int i = 1 ; i <= N ; ++ i)
        for (int j = 1 ; j <= N ; ++ j)
            printf("%lld%c", qwq.M[i][j], " \n"[j == N]) ;
}

四、二分图匹配

#include <cstdio>
#include <cstring>
#include <iostream>

#define R register 
#define MAXN 2000100
#define to(k) E[k].to

using namespace std  ; 
struct Edge{
    int to, next, v ;
}E[MAXN] ; int head[MAXN], Ans, cnt ;
int N, M, Ew, A, B, used[MAXN] ; bool vis[MAXN] ;

inline void Add(int u, int v) {
    if (u < 1 || v < 1 || u > N || v > M) return ;
    E[++ cnt].to = v, E[cnt].next = head[u], head[u] = cnt ;
}
bool path(int u){
    for (int k = head[u] ; k ; k = E[k].next){
        if (!vis[to(k)]){
            vis [to(k)] = 1 ;
            if (!used[to(k)] || path(used[to(k)])){
                used[to(k)] = u ;
                return 1 ;
            }
        }
    }
    return 0 ;
}
int main(){
    cin >> N >> M >> Ew ;
    while(Ew --) scanf("%d%d", &A, &B), Add(A, B) ;
    for (R int i = 1 ; i <= N ; ++ i)       
        memset(vis, 0, sizeof(vis)), Ans += (int)path(i) ;
    cout << Ans << endl ; return 0 ;
}

剩下的回来再整理吧~~

猜你喜欢

转载自www.cnblogs.com/pks-t/p/9938275.html
rm
今日推荐
《美国对全球网络空间安全与发展的威胁和破坏》报告发布
火速冲上 GitHub 热榜 —— 开源编程语言、框架哪有这么可爱？
北京人形机器人创新中心发布全球首个纯电驱拟人奔跑的全尺寸人形机器人“天工”
LFOSSA 源来如此公开课 | 掌握云原生未来：CNCF 认证全面攻略与备考秘籍
周排行
循环神经网络（rnn）讲解
Tigao教程四：单独的关节运动
金蝶K3WISE15.0-注册套打教程
如何在Mac上配置Kubernetes
Android应用结束自身进程的方法
SpringMVC学习 十三 拦截器栈
中国驻洛杉矶总领馆举行新春招待会
HttpClient get post 发送
11 - three.js 笔记 - 绘制三维字体模型
Mysql递归获取某个父节点下面的所有子节点和子节点上的所有父节点
每日归档
更多
2024-05-01(4)
2024-04-30(1)
2024-04-29(40)
2024-04-28(0)
2024-04-27(56)
2024-04-26(39)
2024-04-25(22)
2024-04-24(36)
2024-04-23(26)
2024-04-22(39)

代码天地 © 2018 codetd.com

境外服务器   最新电视剧2022