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。
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<queue>
5 #define RG register
6 using namespace std;
7
8 char s[2000005];
9 int len, nex[2000005], flag[30];
10 int q[2000005];
11
12 int main ( ) {
13 freopen ( "str.in", "r", stdin );
14 freopen ( "str.out", "w", stdout );
15 scanf ( "%s", s );
16 len = strlen ( s );
17 for ( RG int i = 0; i < len; i ++ )
18 flag[s[i]-‘A‘] = 1;
19 int fl = 0;
20 for ( RG int i = 0; i < 26; i ++ )
21 if ( !flag[i] ) {
22 fl = 1; break;
23 }
24 if ( fl ) {
25 printf ( "QwQ" ); return 0;
26 }
27 int num = 0, ans = 0x3f3f3f3f, h = 0, t = 0;
28 memset ( flag, 0, sizeof ( flag ) );
29 for ( RG int i = 0; i < len; i ++ ) {
30 q[++t] = s[i]-‘A‘; flag[s[i]-‘A‘] ++;
31 if ( flag[s[i]-‘A‘] == 1 ) num ++;
32 while ( t-h && num == 26 ) {
33 ans = min ( ans, t-h );
34 int x = q[h+1]; h ++;
35 flag[x] --;
36 if ( !flag[x] ) num --;
37 }
38 }
39 printf ( "%d", ans );
40 }
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。
1 #include<iostream>
2 #include<cstdio>
3 #include<cmath>
4 #define ll long long
5 using namespace std;
6
7 ll x, y;
8 int num1[400001], num2[400001];
9 int prime[400001], tot, isnot[400001];
10
11 void read ( ll &x ) {
12 x = 0; char ch = getchar ( ); int t = 1;
13 while ( ch > ‘9‘ || ch < ‘0‘ ) {
14 if ( ch == ‘-‘ ) t = -1; ch = getchar ( );
15 }
16 while ( ch >= ‘0‘ && ch <= ‘9‘ ) {
17 x = x * 10 + ch - ‘0‘;
18 ch = getchar ( );
19 }
20 x = x * t;
21 }
22
23 inline int gcd ( int a, int b ) {
24 return b == 0 ? a : gcd ( b, a % b );
25 }
26
27 ll erfen ( ll qwq ) {
28 ll l = 1, r = min ( sqrt ( qwq ), 1e6 ), res;
29 while ( l <= r ) {
30 ll mid = ( l + r ) >> 1;
31 if ( mid * mid * mid <= qwq ) {
32 l = mid + 1; res = mid;
33 } else r = mid - 1;
34 }
35 return res;
36 }
37
38 int main ( ) {
39 freopen ( "game.in", "r", stdin );
40 freopen ( "game.out", "w", stdout );
41 int T;
42 scanf ( "%d", &T );
43 while ( T -- ) {
44 int fl = 0;
45 read ( x ); read ( y );
46 ll g = 1ll * x * y;
47 ll qwq = erfen ( g );
48 if ( qwq * qwq * qwq != g ) fl = 1;
49 if ( x % qwq != 0 || y % qwq != 0 ) fl = 1;
50 if ( fl ) printf ( "No\n" );
51 else printf ( "Yes\n" );
52 }
53 return 0;
54 }
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。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<queue>
5 #define inf 0x3f3f3f3f
6 using namespace std;
7
8 int r, c;
9 char a[55][55];
10 int num1[55][55], num2[55][55], cnt1, cnt2, s, t;
11
12 int stot = 1, h[10005], tov[200005], nex[200005], f[200005], hh[10005];
13 void add ( int u, int v, int ff ) {
14 tov[++stot] = v;
15 f[stot] = ff;
16 nex[stot] = h[u];
17 h[u] = stot;
18
19 tov[++stot] = u;
20 f[stot] = 0;
21 nex[stot] = h[v];
22 h[v] = stot;
23 }
24
25 int dep[1005], vis[1005];
26 queue < int > q;
27 bool bfs ( ) {
28 memset ( dep, 0, sizeof ( dep ) );
29 memset ( vis, 0, sizeof ( vis ) );
30 q.push ( s ); vis[s] = 1;
31 while ( !q.empty ( ) ) {
32 int u = q.front ( ); q.pop ( );
33 for ( int i = h[u]; i; i = nex[i] ) {
34 int v = tov[i];
35 if ( !vis[v] && f[i] ) {
36 dep[v] = dep[u] + 1;
37 vis[v] = 1;
38 q.push ( v );
39 }
40 }
41 }
42 return vis[t];
43 }
44
45 int dfs ( int u, int delta ) {
46 if ( u == t ) return delta;
47 int res = 0;
48 for ( int i = hh[u]; i && delta; i = nex[i] ) {
49 int v = tov[i];
50 if ( dep[v] == dep[u] + 1 && f[i] ) {
51 int dd = dfs ( v, min ( f[i], delta ) );
52 f[i] -= dd;
53 f[i^1] += dd;
54 delta -= dd;
55 res += dd;
56 hh[u] = i;
57 }
58 }
59 return res;
60 }
61
62 void debug ( ) {
63 for ( int i = 0; i < r; i ++ ) {
64 for ( int j = 0; j < c; j ++ )
65 printf ( "%d ", num1[i][j] );
66 cout << endl;
67 }
68 cout << endl;
69 for ( int i = 0; i < r; i ++ ) {
70 for ( int j = 0; j < c; j ++ )
71 printf ( "%d ", num2[i][j] );
72 cout << endl;
73 }
74 }
75
76 int main ( ) {
77 freopen ( "cover.in", "r", stdin );
78 freopen ( "cover.out", "w", stdout );
79 scanf ( "%d%d", &r, &c );
80 for ( int i = 0; i < r; i ++ )
81 scanf ( "%s", a[i] );
82 for ( int i = 0; i < r; i ++ )
83 for ( int j = 0; j < c; j ++ ) {
84 if ( j != 0 && a[i][j] == ‘*‘ && a[i][j-1] == ‘*‘ ) num1[i][j] = num1[i][j-1];
85 else if ( a[i][j] == ‘*‘ ){
86 cnt1 ++; num1[i][j] = cnt1;
87 }
88 if ( i != 0 && a[i][j] == ‘*‘ && a[i-1][j] == ‘*‘ ) num2[i][j] = num2[i-1][j];
89 else if ( a[i][j] == ‘*‘ ){
90 cnt2 ++; num2[i][j] = cnt2;
91 }
92 }
93 //debug ( );
94 for ( int i = 0; i < r; i ++ )
95 for ( int j = 0; j < c; j ++ )
96 if ( a[i][j] == ‘*‘ ) {
97 //printf ( "%d->%d+%d\n", num1[i][j], num2[i][j], cnt1 );
98 add ( num1[i][j], num2[i][j] + cnt1, 1 );
99 }
100 s = 0, t = cnt1+cnt2+1;
101 for ( int i = 1; i <= cnt1; i ++ )
102 add ( s, i, 1 );
103 for ( int i = 1; i <= cnt2; i ++ )
104 add ( i + cnt1, t, 1 );
105 int ans = 0;
106 while ( bfs ( ) ) {
107 for ( int i = 0; i <= cnt1 + cnt2 + 1; i ++ )
108 hh[i] = h[i];
109 ans += dfs ( s, 0x3f3f3f3f );
110 }
111 printf ( "%d", ans );
112 return 0;
113 }
原文地址：https://www.cnblogs.com/wans-caesar-02111007/p/9487069.html
时间： 2024-10-11 07:56:32