模板，最后挣扎一下

我不想退役，一定要拿省一

奥利给

1. 堆

#include<bits/stdc++.h>
using namespace std;
#define rint register int
int n;
priority_queue< int, vector< int >, greater< int > > que;
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
int main( void ){
    n = read();
    for( rint i = 1; i <= n; i++ ){
        int temp; temp = read();
        if( temp == 1 ) que.push( read() );
        if( temp == 2 ) printf( "%d\n", que.top() );
        if( temp == 3 ) que.pop();
    }
    return 0;
}

我觉得我的码风还挺好看的吧……
注意答案的换行

2. 负环

竟然是一道蓝题……
跑\(spfa\)，注意使用的是普通队列
更新一条边使用的点数达到\(n\)就说明他自己更新了自己，存在负环
多组数据注意清空\(head\)数组
输出记得粘贴复制，不要手打
注意题目所说连边方式

#include<bits/stdc++.h>
using namespace std;
#define rint register int
int T, n, m, cnt;
int head[2010], dis[2010], cnte[2010], vis[2010];
struct edge{
    int nxt, to, val;
}a[6010];
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
inline void addedge( int x, int y, int z ){
    a[++cnt].nxt = head[x];
    a[cnt].to = y;
    a[cnt].val = z;
    head[x] = cnt;
}
inline bool spfa( int x ){
    memset( dis, 0x3f, sizeof( dis ) );
    memset( cnte, 0, sizeof( cnte ) );
    memset( vis, 0, sizeof( vis ) );
    queue< int > que;
    que.push( x );
    dis[x] = 0;
    vis[x] = 1;
    while( que.size() ){
        int now = que.front();
        que.pop();
        vis[now] = 0;
        if( cnte[now] >= n ) return 0;
        for( rint i = head[now]; i; i = a[i].nxt ){
            int v = a[i].to, w = a[i].val;
            if( dis[v] > dis[now] + w ){
                dis[v] = dis[now] + w;
                if( !vis[v] ){
                    cnte[v]++;
                    vis[v] = 1;
                    que.push( v );
                    if( cnte[v] >= n ) return 0;
                }
            }
        }
    }
    return 1;
}
int main( void ){
    T = read();
    while( T-- ){
        cnt = 0;
        memset( head, 0, sizeof( head ) );
        n = read(); m = read(); 
        for( rint i = 1; i <= m; i++ ){
            int a, b, c; a = read(); b = read(); c = read(); 
            addedge( a, b, c ); if( c >= 0 ) addedge( b, a, c );
        }
        if( spfa( 1 ) ) printf( "N0\n" );
        else printf( "YE5\n" );
    }
    return 0;
}

3. 并查集

疯狂写\(T\) \(22\)分钟
只因为我把\(i <= m\)写成\(i = m\)
草
记得初始化

#include<bits/stdc++.h>
using namespace std;
#define rint register int
int n, m;
int fa[10010];
inline int getf( int now ){
    if( fa[now] == now ) return now;
    else return fa[now] = getf( fa[now] );
} 
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
int main( void ){
    n = read(); m = read();
    for( rint i = 1; i <= n; i++ ) fa[i] = i; 
    for( rint i = 1; i <= m; i++ ){
        int x, y, z; x = read(); y = read(); z = read();
        if( x == 1 ) fa[getf(z)] = getf(y);
        else if( getf(y) == getf(z) ) printf( "Y\n" ); else printf( "N\n" );
    }
    return 0;
}

4. 线性筛

记得是\(j\) % \(prime[i]==0\)

#include<bits/stdc++.h>
using namespace std;
#define rint register int
int n, m, cnt;
int vis[10000010], prime[3000010];
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
int main( void ){
    n = read(); m = read();
    vis[0] = vis[1] = 1;
    for( rint i = 2; i <= n; i++ ){
        if( !vis[i] ) prime[++cnt] = i;
        for( rint j = 1; j <= cnt && prime[j] * i <= n; j++ ){
            vis[i * prime[j]] = 1;
            if( i % prime[j] == 0 ) break ;
        }
    }
    for( rint i = 1; i <= m; i++ ){
        int temp; temp = read();
        if( !vis[temp] ) printf( "Yes\n" ); else printf( "No\n" );
    }
    return 0;
}

5. 快速幂

第三次写错……
下次一定不会\(QAQ\)

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
ll b, p, k;
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
inline ll qpow( int x, int y, int mod ){
    if( y == 0 ) return 1;
    ll re = 1;
    while( y ){
        if( y % 2 ) re = ( re * x ) % mod;
        y >>= 1;
        x = ( x * x ) % mod;
    }
    return re % mod;
}
int main( void ){
    b = read(); p = read(); k = read();
    printf( "%lld^%lld mod %lld=%lld", b, p, k, ( qpow( b % k, p, k ) % k ) );
    return 0;
}

6. 最小生成树

记得写对并查集
记得初始化并查集

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
int n, m, cnt, cnt2;
ll ans;
int fa[5010];
struct edge{
    int u, v, val;
}a[200010];
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
inline bool cmp( edge a, edge b ){
    return a.val < b.val;
}
inline int getf( int now ){
    if( fa[now] == now ) return now; else return fa[now] = getf( fa[now] );
}
int main( void ){
    n = read(); m = read();
    for( rint i = 1; i <= m; i++ ){
        int u, v, w; u = read(); v = read(); w = read(); 
        a[i].u = u, a[i].v = v, a[i].val = w;
    }
    sort( a + 1, a + 1 + m, cmp );
    for( rint i = 1; i <= n; i++ ) fa[i] = i;
    for( rint i = 1; i <= m; i++ ){
        if( cnt2 == n - 1 ) break ;
        if( getf( a[i].u ) == getf( a[i].v ) ) continue ;
        else {
            cnt2++;
            fa[getf( a[i].v )] = fa[a[i].u];
            ans += a[i].val;
        }
    }
    if( cnt2 == n-1 ) printf( "%lld", ans );
    else printf( "orz" );
    return 0;
}

7. 线段树

为什么先打的这个？？？
跟\(YKY\)竞速输了
(就差一点……

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
int n, m;
ll a[1000010];
struct node{
    int left, right;
    ll pre, tag;
}tree[4000010];
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' && ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
inline void build( ll now, ll l, ll r ){
    tree[now].left = l, tree[now].right = r;
    if( l == r ){
        tree[now].pre = a[l];
        return ;
    }
    ll mid = l + r >> 1;
    build( now << 1, l, mid ); build( now << 1 | 1, mid + 1, r );
    tree[now].pre = tree[now << 1].pre + tree[now << 1 | 1].pre;
}
inline void spread( ll now ){
    if( tree[now].tag ){
        tree[now << 1].pre += tree[now].tag * ( tree[now << 1].right - tree[now << 1].left + 1 );
        tree[now << 1 | 1].pre += tree[now].tag * ( tree[now << 1 | 1].right - tree[now << 1 | 1].left + 1 );
        tree[now << 1].tag += tree[now].tag;
        tree[now << 1 | 1].tag += tree[now].tag;
        tree[now].tag = 0;
    }
}
inline void update( ll now, ll l, ll r, ll val ){
    if( l <= tree[now].left && r >= tree[now].right ){
        tree[now].pre += ( tree[now].right - tree[now].left + 1 ) * val;
        tree[now].tag += val;
        return ;
    }
    spread( now );
    ll mid = tree[now].left + tree[now].right >> 1;
    if( mid < r ) update( now << 1 | 1, l, r, val );
    if( l <= mid ) update( now << 1, l, r, val );
    tree[now].pre = tree[now << 1].pre + tree[now << 1 | 1].pre;
}
inline ll query( int now, int l, int r ){
    ll ans = 0;
    if( l <= tree[now].left && r >= tree[now].right ){
        return tree[now].pre;
    }
    spread( now );
    ll mid = tree[now].left + tree[now].right >> 1;
    if( mid < r ) ans += query( now << 1 | 1, l, r );
    if( l <= mid ) ans += query( now << 1, l, r );
    return ans;
}
int main( void ){
    n = read(); m = read();
    for( rint i = 1; i <= n; i++ ){
        a[i] = read();
    }
    build( 1, 1, n );
    for( rint i = 1; i <= m; i++ ){
        int bok, x, y, k;
        bok = read();
        if( bok == 1 ){
            x = read(); y = read(); k = read();
            update( 1, x, y, k );
        } else {
            x = read(); y = read();
            printf( "%lld\n", query( 1, x, y ) );
        }
    }
    return 0;
}

8. lca

为什么又打的这个？
因为我竞速又输了……
\(8\)说了，自闭去了

#include<bits/stdc++.h>
using namespace std;
#define rint register int
int n, m, s, cnt;
struct edge{
    int nxt, to;
}a[1000010];
int fa[500010], top[500010], siz[500010], dep[500010], son[500010], head[500010];
inline int read( void ){
    int re = 0, f = 1; char ch = getchar();
    while( ch > '9' || ch < '0' ){
        if( ch == '-' ) f = -1;
        ch = getchar();
    }
    while( ch >= '0' &&  ch <= '9' ){
        re = re * 10 + ch - '0';
        ch = getchar();
    }
    return re * f;
}
inline void addedge( int x, int y ){
    a[++cnt].to = y;
    a[cnt].nxt = head[x];
    head[x] = cnt;
}
inline void dfs( int now, int f, int d ){
    dep[now] = d; fa[now] = f; siz[now] = 1;
    for( rint i = head[now]; i; i = a[i].nxt ){
        int v = a[i].to;
        if( v == f ) continue ;
        dfs( v, now, d + 1 );
        siz[now] += siz[v];
        if( siz[v] > siz[son[now]] ) son[now] = v;
    }
    return ;
}
inline void dfs2( int now, int topf ){
    top[now] = topf;
    if( !son[now] ) return ;
    dfs2( son[now], topf );
    for( rint i = head[now]; i; i = a[i].nxt ){
        int v = a[i].to;
        if( v == fa[now] || v == son[now] ) continue ;
        dfs2( v, v );
    }
    return ;
}
inline int lca( int x, int y ){
    while( top[x] != top[y] ){
        if( dep[top[x]] < dep[top[y]] ) swap( x, y );
        x = fa[top[x]];
    }
    if( dep[x] > dep[y] ) return y; else return x;
}
int main( void ){
    n = read(); m = read(); s = read();
    for( rint i = 1; i < n; i++ ){
        int u, v; u = read(); v = read();
        addedge( u, v ); addedge( v, u );
    }
    dfs( s, s, 1 );
    dfs2( s, s );
    for( rint i = 1; i <= m; i++ ){
        int u, v; u = read(); v = read();
        printf( "%d\n", lca( u, v ) );
    }
    return 0;
}

9.