【8.16校内测试】【队列】【数学】【网络流/二分图最大匹配】

在一个队列中一次加入每一个字符，每次更新当前队列中的状态，当满足存在26个不同字符时，更新答案，删除队首。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define RG register
using namespace std;

char s[2000005];
int len, nex[2000005], flag[30];
int q[2000005];

int main ( ) {
    freopen ( "str.in", "r", stdin );
    freopen ( "str.out", "w", stdout );
    scanf ( "%s", s );
    len = strlen ( s );
    for ( RG int i = 0; i < len; i ++ )
        flag[s[i]-'A'] = 1;
    int fl = 0;
    for ( RG int i = 0; i < 26; i ++ )    
        if ( !flag[i] ) {
            fl = 1; break;
        }
    if ( fl ) {
        printf ( "QwQ" );     return 0;
    }
    int num = 0, ans = 0x3f3f3f3f, h = 0, t = 0;
    memset ( flag, 0, sizeof ( flag ) );
    for ( RG int i = 0; i < len; i ++ ) {
        q[++t] = s[i]-'A'; flag[s[i]-'A'] ++;
        if ( flag[s[i]-'A'] == 1 ) num ++;
        while ( t-h && num == 26 ) {
            ans = min ( ans, t-h );
            int x = q[h+1]; h ++;
            flag[x] --;
            if ( !flag[x] ) num --;
        }
    }
    printf ( "%d", ans );
}

一开始想的分解质因数，再通过每个质因子的个数来判断是否成立，可是一开始就错了...以为1e9开方是1e3...

方法是先将x和y乘起来，因为题目有一个性质，他们的乘积一定是一个数的3次方，设这个数为k，因为x和y中每次游戏要不是有一个k1值，要不是有两个，所以x和y必然可以整除k。三个判断条件即可。【注意】二分求k值时不能让k大于1e6,三方爆long long。

#include<iostream>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;

ll x, y;
int num1[400001], num2[400001];
int prime[400001], tot, isnot[400001];

void read ( ll &x ) {
    x = 0; char ch = getchar ( ); int t = 1;
    while ( ch > '9' || ch < '0' ) {
        if ( ch == '-' ) t = -1; ch = getchar ( );
    }
    while ( ch >= '0' && ch <= '9' ) {
        x = x * 10 + ch - '0';
        ch = getchar ( );
    }
    x = x * t;
}

inline int gcd ( int a, int b ) {
    return b == 0 ? a : gcd ( b, a % b );
}

ll erfen ( ll qwq ) {
    ll l = 1, r = min ( sqrt ( qwq ), 1e6 ), res;
    while ( l <= r ) {
        ll mid = ( l + r ) >> 1;
        if ( mid * mid * mid <= qwq ) {
            l = mid + 1; res = mid;
        } else r = mid - 1;
    }
    return res;
}

int main ( ) {
    freopen ( "game.in", "r", stdin );
    freopen ( "game.out", "w", stdout );
    int T;
    scanf ( "%d", &T );
    while ( T -- ) {
        int fl = 0;
        read ( x ); read ( y );
        ll g = 1ll * x * y;
        ll qwq = erfen ( g );
        if ( qwq * qwq * qwq != g ) fl = 1;
        if ( x % qwq != 0 || y % qwq != 0 ) fl = 1;
        if ( fl ) printf ( "No
" );
        else printf ( "Yes
" );
    }
    return 0;
}

比较经典的一道题，分别把按行放木板和按列放木板给每一块泥地标号，可以连着放的号数一样。把每一个泥地的行号连向列号，跑最小割或者最大匹配即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define inf 0x3f3f3f3f
using namespace std;

int r, c;
char a[55][55];
int num1[55][55], num2[55][55], cnt1, cnt2, s, t;

int stot = 1, h[10005], tov[200005], nex[200005], f[200005], hh[10005];
void add ( int u, int v, int ff ) {
    tov[++stot] = v;
    f[stot] = ff;
    nex[stot] = h[u];
    h[u] = stot;
    
    tov[++stot] = u;
    f[stot] = 0;
    nex[stot] = h[v];
    h[v] = stot;
}

int dep[1005], vis[1005];
queue < int > q;
bool bfs ( ) {
    memset ( dep, 0, sizeof ( dep ) );
    memset ( vis, 0, sizeof ( vis ) );
    q.push ( s );    vis[s] = 1;
    while ( !q.empty ( ) ) {
        int u = q.front ( ); q.pop ( );
        for ( int i = h[u]; i; i = nex[i] ) {
            int v = tov[i];
            if ( !vis[v] && f[i] ) {
                dep[v] = dep[u] + 1;
                vis[v] = 1;
                q.push ( v );
            }
        }
    }
    return vis[t];
}

int dfs ( int u, int delta ) {
    if ( u == t ) return delta;
    int res = 0;
    for ( int i = hh[u]; i && delta; i = nex[i] ) {
        int v = tov[i];
        if ( dep[v] == dep[u] + 1 && f[i] ) {
            int dd = dfs ( v, min ( f[i], delta ) );
            f[i] -= dd;
            f[i^1] += dd;
            delta -= dd;
            res += dd;
            hh[u] = i;
        }
    }
    return res;
}

void debug ( ) {
    for ( int i = 0; i < r; i ++ ) {
        for ( int j = 0; j < c; j ++ )
            printf ( "%d ", num1[i][j] );
        cout << endl;
    }
    cout << endl;
    for ( int i = 0; i < r; i ++ ) {
        for ( int j = 0; j < c; j ++ )
            printf ( "%d ", num2[i][j] );
        cout << endl;
    }
}

int main ( ) {
    freopen ( "cover.in", "r", stdin );
    freopen ( "cover.out", "w", stdout );
    scanf ( "%d%d", &r, &c );
    for ( int i = 0; i < r; i ++ )
        scanf ( "%s", a[i] );
    for ( int i = 0; i < r; i ++ )
        for ( int j = 0; j < c; j ++ ) {
            if ( j != 0 && a[i][j] == '*' && a[i][j-1] == '*' )     num1[i][j] = num1[i][j-1];
            else if ( a[i][j] == '*' ){
                cnt1 ++;    num1[i][j] = cnt1;
            }
            if ( i != 0 && a[i][j] == '*' && a[i-1][j] == '*' )     num2[i][j] = num2[i-1][j];
            else if ( a[i][j] == '*' ){
                cnt2 ++;     num2[i][j] = cnt2;
            }
        }
    //debug ( );
    for ( int i = 0; i < r; i ++ )
        for ( int j = 0; j < c; j ++ )
            if ( a[i][j] == '*' ) {
                //printf ( "%d->%d+%d
", num1[i][j], num2[i][j], cnt1 );
                add ( num1[i][j], num2[i][j] + cnt1, 1 );
            }
    s = 0, t = cnt1+cnt2+1;
    for ( int i = 1; i <= cnt1; i ++ )
        add ( s, i, 1 );
    for ( int i = 1; i <= cnt2; i ++ )
        add ( i + cnt1, t, 1 );
    int ans = 0;
    while ( bfs ( ) ) {
        for ( int i = 0; i <= cnt1 + cnt2 + 1; i ++ )
            hh[i] = h[i];
        ans += dfs ( s, 0x3f3f3f3f );
    }
    printf ( "%d", ans );
    return 0;
}