A. Salem and Sticks
签.
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 1010
5 int n, a[N];
6
7 int work(int x)
8 {
9 int res = 0;
10 for (int i = 1; i <= n; ++i)
11 res += max(0, abs(x - a[i]) - 1);
12 return res;
13 }
14
15 int main()
16 {
17 while (scanf("%d", &n) != EOF)
18 {
19 for (int i = 1; i <= n; ++i) scanf("%d", a + i);
20 int Min = (int)1e9, pos = -1;
21 for (int i = 1; i <= 100; ++i)
22 {
23 int tmp = work(i);
24 if (tmp < Min)
25 {
26 Min = tmp;
27 pos = i;
28 }
29 }
30 printf("%d %d\n", pos, Min);
31 }
32 return 0;
33 }
B. Zuhair and Strings
签.
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 200010
5 int n, k;
6 char s[N];
7
8 int work(char c)
9 {
10 int res = 0;
11 int tmp = 0;
12 for (int i = 1; i <= n; ++i)
13 {
14 if (s[i] != c) tmp = 0;
15 else
16 {
17 ++tmp;
18 if (tmp == k)
19 {
20 ++res;
21 tmp = 0;
22 }
23 }
24 }
25 return res;
26 }
27
28 int main()
29 {
30 while (scanf("%d%d", &n, &k) != EOF)
31 {
32 scanf("%s", s + 1);
33 int res = 0;
34 for (int i = ‘a‘; i <= ‘z‘; ++i)
35 res = max(res, work(i));
36 printf("%d\n", res);
37 }
38 return 0;
39 }
C. Ayoub and Lost Array
签.
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define ll long long
5 #define N 200010
6 const ll MOD = (ll)1e9 + 7;
7 int n, l, r;
8 ll a[3], f[N][3];
9
10 int main()
11 {
12 while (scanf("%d%d%d", &n, &l, &r) != EOF)
13 {
14 memset(a, 0, sizeof a);
15 while (l % 3 && l <= r)
16 {
17 a[l % 3]++;
18 ++l;
19 }
20 while (r % 3 && l <= r)
21 {
22 a[r % 3]++;
23 --r;
24 }
25 if (l <= r)
26 {
27 ++a[0];
28 int tmp = (r - l) / 3;
29 a[0] += tmp;
30 a[1] += tmp;
31 a[2] += tmp;
32 }
33 memset(f, 0, sizeof f);
34 f[0][0] = 1;
35 for (int i = 1; i <= n; ++i)
36 {
37 f[i][0] = (f[i - 1][0] * a[0] % MOD + f[i - 1][1] * a[2] % MOD + f[i - 1][2] * a[1] % MOD) % MOD;
38 f[i][1] = (f[i - 1][0] * a[1] % MOD + f[i - 1][1] * a[0] % MOD + f[i - 1][2] * a[2] % MOD) % MOD;
39 f[i][2] = (f[i - 1][0] * a[2] % MOD + f[i - 1][1] * a[1] % MOD + f[i - 1][2] * a[0] % MOD) % MOD;
40 }
41 printf("%lld\n", f[n][0]);
42 }
43 return 0;
44 }
D. Kilani and the Game
签.
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 1010
5 int n, m, p;
6 char G[N][N];
7 int s[10];
8 struct node
9 {
10 int x, y, step;
11 node () {}
12 node (int x, int y, int step) : x(x), y(y), step(step) {}
13 };
14 queue <node> q[10];
15 bool stop()
16 {
17 for (int i = 1; i <= p; ++i) if (!q[i].empty())
18 return false;
19 return true;
20 }
21
22 int Move[][2] =
23 {
24 -1, 0,
25 1, 0,
26 0,-1,
27 0, 1,
28 };
29 bool ok(int x, int y)
30 {
31 if (x < 1 || x > n || y < 1 || y > m || G[x][y] != ‘.‘) return false;
32 return true;
33 }
34
35 void BFS(int id, int cnt)
36 {
37 while (!q[id].empty())
38 {
39 int x = q[id].front().x;
40 int y = q[id].front().y;
41 int step = q[id].front().step;
42 //printf("%d %d %d %d\n", x, y, id, step);
43 if (step / s[id] >= cnt) return;
44 q[id].pop();
45 for (int i = 0; i < 4; ++i)
46 {
47 int nx = x + Move[i][0];
48 int ny = y + Move[i][1];
49 if (ok(nx, ny))
50 {
51 G[nx][ny] = id + ‘0‘;
52 q[id].push(node(nx, ny, step + 1));
53 }
54 }
55 }
56 }
57
58 int main()
59 {
60 while (scanf("%d%d%d", &n, &m, &p) != EOF)
61 {
62 for (int i = 1; i <= p; ++i) scanf("%d", s + i);
63 for (int i = 1; i <= n; ++i) scanf("%s", G[i] + 1);
64 for (int i = 1; i <= p; ++i) while (!q[i].empty()) q[i].pop();
65 for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) if (isdigit(G[i][j]))
66 q[G[i][j] - ‘0‘].push(node(i, j, 0));
67 int cnt = 1;
68 while (1)
69 {
70 for (int i = 1; i <= p; ++i) BFS(i, cnt);
71 ++cnt;
72 if (stop()) break;
73 }
74 int ans[10];
75 memset(ans, 0, sizeof ans);
76 for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) if (isdigit(G[i][j]))
77 ++ans[G[i][j] - ‘0‘];
78 //for (int i = 1; i <= n; ++i) printf("%s\n", G[i] + 1);
79 for (int i = 1; i <= p; ++i) printf("%d%c", ans[i], " \n"[i == p]);
80 }
81 return 0;
82 }
E. Helping Hiasat
Upsolved.
题意：
两种操作
- 更改自己的handle
- 伙伴查询handle
如果一个伙伴在每次查询时显示的都是自己名字，那么他就会开心
问 最多可以让多少人开心
思路：
法一：
一张图的最大独立集是选出一个点集，使得任意两点不相邻
一张图的最大团是选出一个点集，使得任意两点之间有边相连
一张无向图的补图的最大图就是原图的最大独立集
我们发现这道题两个1之间的所有点都是不能一起happy的，那我们给他们两两之间连上边
然后求补图的最大团即可
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 110
5 int n, m, t;
6 int g[N][N];
7 int dp[N];
8 int stk[N][N];
9 int mx;
10
11 map <string, int> mp;
12 int get(string s)
13 {
14 if (mp.find(s) != mp.end()) return mp[s];
15 else mp[s] = t++;
16 return mp[s];
17 }
18
19 int DFS(int n, int ns, int dep)
20 {
21 if (ns == 0)
22 {
23 mx = max(mx, dep);
24 return 1;
25 }
26 int i, j, k, p, cnt;
27 for (i = 0; i < ns; ++i)
28 {
29 k = stk[dep][i];
30 cnt = 0;
31 if (dep + n - k <= mx)
32 return 0;
33 if (dep + dp[k] <= mx)
34 return 0;
35 for (j = i + 1; j < ns; ++j)
36 {
37 p = stk[dep][j];
38 if (g[k][p])
39 stk[dep + 1][cnt++] = p;
40 }
41 DFS(n, cnt, dep + 1);
42 }
43 return 1;
44 }
45
46 int clique(int n)
47 {
48 int i, j, ns;
49 for (mx = 0, i = n - 1; i >= 0; --i)
50 {
51 for (ns = 0, j = i + 1; j < n; ++j)
52 {
53 if (g[i][j])
54 stk[1][ns++] = j;
55 }
56 DFS(n, ns, 1);
57 dp[i] = mx;
58 }
59 return mx;
60 }
61
62 vector <int> vec;
63 void add()
64 {
65 vec.erase(unique(vec.begin(), vec.end()), vec.end());
66 for (auto u : vec) for (auto v : vec)
67 g[u][v] = g[v][u] = 0;
68 vec.clear();
69 }
70
71 int main()
72 {
73 while (scanf("%d%d", &n, &m) != EOF)
74 {
75 t = 0;
76 mp.clear();
77 for (int i = 0; i < m; ++i) for (int j = 0; j < m; ++j) g[i][j] = 1;
78 []()
79 {
80 int op; char s[50];
81 for (int nn = 1; nn <= n; ++nn)
82 {
83 scanf("%d", &op);
84 if (op == 1)
85 add();
86 else
87 {
88 scanf("%s", s + 1);
89 vec.push_back(get(s + 1));
90 }
91 }
92 }();
93 add();
94 for (int i = 0; i < m; ++i) g[i][i] = 1;
95 //for (int i = 1; i <= m; ++i) for (int j = 1; j <= m; ++j) printf("%d %d %d\n", i, j, g[i][j]);
96 printf("%d\n", clique(m));
97 }
98 return 0;
99 }
法二：
$m = 40, 可以折半状压，再合起来$
考虑妆压的时候要从子集转移到超集，可以通过$dp上去$
$vp的时候想到折半状压，但是当时是枚举子集来转移，复杂度大大增加..$
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define N 50
5 #define M 1100010
6 int n, m, t;
7 int G[N][N];
8
9 map <string, int> mp;
10 int get(string s)
11 {
12 if (mp.find(s) != mp.end()) return mp[s];
13 else mp[s] = ++t;
14 return mp[s];
15 }
16
17 vector <int> vec;
18 void add()
19 {
20 vec.erase(unique(vec.begin(), vec.end()), vec.end());
21 for (auto u : vec) for (auto v : vec)
22 G[u][v] = 1;
23 vec.clear();
24 }
25
26 int f[M], g[M];
27 int main()
28 {
29 while (scanf("%d%d", &n, &m) != EOF)
30 {
31 t = 0; mp.clear();
32 memset(G, 0, sizeof G);
33 int op; char s[N];
34 for (int nn = 1; nn <= n; ++nn)
35 {
36 scanf("%d", &op);
37 if (op == 1) add();
38 else
39 {
40 scanf("%s", s + 1);
41 vec.push_back(get(s + 1));
42 }
43 }
44 add();
45 for (int i = 1; i <= m; ++i)
46 G[i][i] = 0;
47 int s1 = m / 2, s2 = m - s1;
48 for (int i = 1; i < (1 << s1); i <<= 1) f[i] = 1;
49 for (int i = 1; i < (1 << s2); i <<= 1) g[i] = 1;
50 f[0] = g[0] = 0;
51 for (int i = 1; i < (1 << s1); ++i)
52 {
53 for (int j = 0; j < s1; ++j) if (!((i >> j) & 1))
54 {
55 int flag = 1;
56 for (int k = 0; k < s1; ++k) if (((i >> k) & 1) && G[k + 1][j + 1])
57 {
58 flag = 0;
59 break;
60 }
61 f[i | (1 << j)] = max(f[i | (1 << j)], f[i] + flag);
62 }
63 }
64 for (int i = 1; i < (1 << s2); ++i)
65 {
66 for (int j = 0; j < s2; ++j) if (!((i >> j) & 1))
67 {
68 int flag = 1;
69 for (int k = 0; k < s2; ++k) if (((i >> k) & 1) && G[s1 + k + 1][s1 + j + 1])
70 {
71 flag = 0;
72 break;
73 }
74 g[i | (1 << j)] = max(g[i | (1 << j)], g[i] + flag);
75 }
76 }
77 int res = 0;
78 for (int i = 0; i < (1 << s1); ++i)
79 {
80 int s3 = (1 << s2) - 1;
81 for (int j = 0; j < s1; ++j) if ((i >> j) & 1)
82 {
83 for (int k = 0; k < s2; ++k) if (G[j + 1][s1 + k + 1] && ((s3 >> k) & 1))
84 s3 ^= (1 << k);
85 }
86 res = max(res, f[i] + g[s3]);
87 }
88 printf("%d\n", res);
89 }
90 return 0;
91 }
原文地址：https://www.cnblogs.com/Dup4/p/10342214.html
时间： 2024-10-08 11:00:19