自用板子

Posted 2022-12-21 phemiku

最短路

单源最短路（1~n）

//SPFA算法

#include<bits/stdc++.h>
using namespace std;
const int N= 15000;
struct bian
    int x,y,d,next;
;
int last[N],len,d[N];
int List[N],tail,head,v[N];
bian a[210000];

void ins(int x,int y,int d)

  len++;
  a[len].x=x;a[len].y=y;a[len].d=d;
  a[len].next=last[x];
  last[x]=len;    


int main()

  int n,m,x,y,c,st,ed;
  scanf("%d%d",&n,&m); //点，边 
  for(int i=1;i<=m;i++)
      scanf("%d%d%d",&x,&y,&c);
      ins(x,y,c);
      ins(y,x,c);
  
  for(int i=1;i<=n;i++)d[i]=9999999999;
  st=1;ed=n;
  d[st]=0;v[st]=1;
  List[1]=st;head=1;tail=2;
  while(head!=tail)
      x=List[head];
      for(int k=last[x];k;k=a[k].next)
          y=a[k].y;
          if(d[y]>d[x]+a[k].d)
              d[y]=d[x]+a[k].d;
              
              if(v[y]==0)
                  v[y]==1;
                  List[tail]=y;
                  tail++;
                  if(tail==n+1) tail=1;
              
          
      
      List[head]=0;
      head++;
      if(head==n+1)head=1;
      v[x]=0; 
  
  printf("%d",d[n]);

//Dijsktra裸题 忘了题号了
#include<iostream>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f 
const int N = 150;
int n,m,x,y,z,p;
int a[N][N],dis[N];
bool vis[N];
void dij(int x, int y)

    for(int i = 1; i <= n; i++)  
      dis[i] = a[1][i];  //单源最短路 
      vis[i] = false; 
    
    vis[x] = 1;
    
    for(int i = 1; i <= n; i++)
        int minn = INF;
      for(int j = 1; j <= n; j++)
          if(!vis[j]&&dis[j]<minn)
              minn = dis[j];
              p = j;
          
      
      vis[p] = 1;
      for(int j = 1; j <= n; j++)
          if(!vis[j]&&dis[p]+a[p][j]<dis[j])
           dis[j] = dis[p] + a[p][j];
            
    

int main()

  while(cin >> n >> m)
      if(n == 0 && m == 0)  break;
      memset(a,INF,sizeof(a));
    while(m--)
        cin >> x >> y >> z;
        a[x][y] = a[y][x] = z;
    
    dij(1,n);
    cout << dis[n] << endl;      
  
  return 0;

 

 2.快速幂和快速乘

#include<bits/stdc++.h>
using namespace std;
#define int long long    //这样就不用每次开long long了
const int N = 100050;
inline int read()      //快读
    char c = getchar();
    int x = 0, f = 1;
    while(c < ‘0‘ || c > ‘9‘) 
        if(c == ‘-‘) f = -1;
        c = getchar();
    
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * f;

inline int qpow(int a,int b,int mod)  //快速幂
    for(int c = 1; ;a = a * a % mod)
        if(b & 1)c = c * a % mod;
        if(!(b >>= 1))return c;
    


inline int qmul(int a,int b,int mod)   //快速乘
    for(int c = 0; ;a = (a << 1) % mod)
        if(b & 1)c = (c + a) % mod;
        if(!(b >>= 1))return c;
    


int n,m,a,b,q,c,d,r;
signed t;
signed main()

    t = read();
   for(int i = 1;i <= t; i++)
     a = read();b = read();q = read();
     c = read();d = read();r = read();
     cout << qpow(a,b,q) << " "<< qmul(c,d,r) << endl ;
   
   

 3.二分

int l = 1;
    int r = d;
    while (l <= r)
        int mid = (l + r) / 2;
        if (check(mid))
            ans = mid;
            l = mid + 1;
        
        else r = mid - 1;
    

 4.高精度

 高精度减法

#include<bits/stdc++.h>
using namespace std;
const int N= 10500;
int na[N],nb[N],nc[N];
bool bj;
string a,b; 
int main()
 
  cin>>a>>b;
  if(a.size()<b.size()||(a.size()==b.size()&&a<b)) //我果然年少又无知…string可以直接比较大小的…（ASCII码 
      bj=1;
      swap(a,b);
  
for(int i=a.size();i>=1;i--)na[i]=a[a.size()-i]-‘0‘;
for(int i=b.size();i>=1;i--)nb[i]=b[b.size()-i]-‘0‘;
  
  int n=max(a.size(),b.size());
  for(int i = 1; i <= n; i ++)
        if(na[i] < nb[i])
            na[i + 1] --;
            na[i] += 10;
        
        nc[i] = na[i] - nb[i];
    
  while(nc[n]==0)n--;
  if(bj)cout<<"-";
  for(int i=n;i>0;i--)cout<<nc[i];
  if(n<1)cout<<"0";

高精度乘法

#include<iostream>
#include<cstring>
#define maxn 100000+10
using namespace std;
int na[maxn],nb[maxn],nc[maxn];
char a[maxn],b[maxn];
void mul()

    int i,j,lena,lenb;
    memset(na,0,sizeof(na));
    memset(nb,0,sizeof(nb));
    memset(nc,0,sizeof(nc));
    lena=strlen(a);
    lenb=strlen(b);
    for(i=0;i<lena;i++) na[i]=a[lena-i-1]-‘0‘;
    for(i=0;i<lenb;i++) nb[i]=b[lenb-i-1]-‘0‘;
    for(i=0;i<lena;i++)
      for(j=0;j<lenb;j++) nc[i+j]+=na[i]*nb[j];
    int max=lena+lenb;
    for(i=0;i<max;i++) nc[i+1]+=nc[i]/10,nc[i]%=10;
    while(!nc[--max]);
    max++;
    for(i=0;i<max;i++) a[i]=nc[max-i-1]+‘0‘;
    a[max]=‘\0‘;

int main()

    while(cin>>a>>b)
    
        mul();
        puts(a);
    
    return 0;

 5.欧几里得（辗转相除）

ll gcd(ll a, ll b)  return !b ? a : gcd(b, a%b); 

6.左偏树

（P3377 【模板】左偏树（可并堆））

 

 

100.乱七八糟的东西

//重载运算符比较大小
struct node   int len;   
   bool operator <(const node &a)const //重载<操作符。可以对两个node使用<操作符进行比较
       return len<a.len;
   
;