A. Lottery
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27 const int maxn = 32;
28 bool vis[maxn+1] = {0};
29
30 int main()
31 {
32 int a = 10, b = 6;
33 for (int i = 0; i < a; ++i)
34 {
35 int x;
36 scanf("%d", &x);
37 vis[x] = true;
38 }
39 int T;
40 scanf("%d", &T);
41 for (int kase = 1; kase <= T; ++kase)
42 {
43 int cnt = 0;
44 for (int i = 0; i < b; ++i)
45 {
46 int x;
47 scanf("%d", &x);
48 cnt += vis[x];
49 }
50 puts(cnt>=3?"Lucky":"Unlucky");
51 }
52 }
B. Permutation Recovery
题意：对于一个n*n的permutation，n<=100，可以看作一个n*n的方阵。现将每一行看作一个长度为n的数列，每一列同理，n行和n列放在一起，打乱顺序。让你找出一组可能的permutation。保证输入有解。
观察：permutation第一个元素会作为开头出现两次，其他元素至多作为开头出现一次。所以找到开头出现两次的元素，把其中一个数列当作第一列，然后将每一列开头元素对应的数列输出即可。
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27 const int maxn = 100+1;
28 int n;
29 vector<int> g[maxn*maxn];
30 vector<int> tar;
31 int main()
32 {
33 scanf("%d", &n);
34 for (int i = 1; i <= 2*n; ++i)
35 {
36 vector<int> tmp(n);
37 for (auto &e : tmp)
38 scanf("%d", &e);
39 if (g[tmp.front()].empty())
40 g[tmp.front()] = tmp;
41 else
42 tar = tmp;
43 }
44 bool first = true;
45 for (auto e : tar)
46 {
47 const auto &v = g[e];
48 for (auto e : v)
49 {
50 if (first)
51 first = false;
52 else
53 printf(" ");
54 printf("%d", e);
55 }
56 }
57 if (!first)
58 puts("");
59 }
WA：数组开小了。
C. Beautiful Tables
题意：给你一个n，不超过1000，让你在n*n的方阵中填1-n的数字，使得每一行的元素和 and 每一列的元素和都能被n整除。问你有多少种填的方案，答案mod 1e9+7。
观察：对于左上角(n-1)*(n-1)的子方阵，随意填数，剩下最后一行和最后一列的数字就会被唯一确定，而且有解。所以答案就是n^[(n-1)*(n-1)]。
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27
28 /*
29 by skydog
30 */
31 #include <iostream>
32 #include <cstdio>
33 #include <vector>
34 #include <utility>
35 #include <algorithm>
36 #include <cmath>
37 #include <cstring>
38 #include <map>
39 #include <set>
40 #include <stack>
41 #include <queue>
42 #include <deque>
43 #include <cassert>
44 #include <list>
45 using namespace std;
46 typedef long long ll;
47 typedef pair<int, int> ii;
48 typedef pair<ll, ll> l4;
49
50 #define mp make_pair
51 #define pb push_back
52 #define db(x) cerr << #x << " = " << x << endl
53
54 const int maxn = 3e5+1;
55 ll len[maxn];
56 int maxl = 88;
57 int n, q;
58 void solve(int l, int r)
59 {
60 vector<l4> v;
61 for (int i = l; i <= min(r, l+maxl-1); ++i)
62 v.push_back(mp(len[i], i));
63 sort(v.begin(), v.end());
64 for (int i = 2; i < v.size(); ++i)
65 if (v[i-2].first + v[i-1].first > v[i].first)
66 {
67 printf("%lld %lld %lld\n", v[i-2].second, v[i-1].second, v[i].second);
68 return;
69 }
70 puts("-1");
71 }
72 const ll inf = 1000000000000000000;
73 int get_maxl()
74 {
75 ll a = 1, b = 1;
76 int cnt = 2;
77 while (b <= inf)
78 {
79 a += b;
80 swap(a, b);
81 ++cnt;
82 }
83 return cnt;
84 }
85 int main()
86 {
87 cerr << "maxl = " << (maxl = get_maxl()) << endl;
88 scanf("%d %d", &n, &q);
89 for (int i = 1; i <= n; ++i)
90 scanf("%lld", len+i);
91 for (int i = 1; i <= q; ++i)
92 {
93 int l, r;
94 scanf("%d %d", &l, &r);
95 solve(l, r);
96 }
97
98 }
D. Triangle Construction
题意：给你n (n <= 3e5) 个长度不超过1e18的木棍，长度计作L[1-n]。有q (q <= 3e5) 组询问，每组询问形如(l, r), 1 <= l <= r <= n。让你找出[l, r]中三个下标不同的木棍，使得这三个木根可以构成一个三角形。如果找不到输出-1。
观察：三条木棍能构成三角形，当且仅当最长边长度严格小于其余两条边长度和。可以想到，如果有一些木棍，找出一组可行解很简单，只需将木棍按照长度排序，然后从小到大找连续的三根木棍检查是否符合条件即可。如果找不到，便无解。
为什么这样是对的。可以考虑枚举最长边，那么为了尽可能达到条件，剩下较短的两条边要选尽量长的，所以只需要考虑相邻的三根木棍就好了。
下面考虑一下无解的情况，把长度排序，最坏的情况肯定是 1, 1, 2, 3, 5, 8, ....，即Fibonacci 数列。但是Fibonacci数列使呈指数级别增长的，算一下发现f[1] = f[2] = 1的Fibonacci数列第88项刚好超过1e18。所以如果木棍个数不小于88，那么一定可以找到解。
所以对于每个询问就只用考虑min(88, r-l+1)个元素，排序后检查是否有解。这样不用预处理，令len = min(88, r-l+1)，单次询问时间为O(len*log(len))。
当然也可以先对每个位置l，暴力计算出使得[l, r]有解的最小的r，记作ans[l]。令len = min(88, n-l+1)，这样处理一个l的复杂度是O(len*log(len))，整体处理所有l的话，令len = min(88, n)，可以做到O(n*len*log(len))。单次询问O(1)，例如询问[l, r]只需要看r和ans[l]的关系即可。如果r < ans[l]，无解，否则直接输出答案即可。
考虑了一下是否可以将上面这种方法利用滑动窗口之类的优化，但是好像不能避开最坏情况。
code：
无预处理，单次询问时间为O(len*log(len))。
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27
28 /*
29 by skydog
30 */
31 #include <iostream>
32 #include <cstdio>
33 #include <vector>
34 #include <utility>
35 #include <algorithm>
36 #include <cmath>
37 #include <cstring>
38 #include <map>
39 #include <set>
40 #include <stack>
41 #include <queue>
42 #include <deque>
43 #include <cassert>
44 #include <list>
45 using namespace std;
46 typedef long long ll;
47 typedef pair<int, int> ii;
48 typedef pair<ll, ll> l4;
49
50 #define mp make_pair
51 #define pb push_back
52 #define db(x) cerr << #x << " = " << x << endl
53
54 const int maxn = 3e5+1;
55 ll len[maxn];
56 int maxl = 88;
57 int n, q;
58 void solve(int l, int r)
59 {
60 vector<l4> v;
61 for (int i = l; i <= min(r, l+maxl-1); ++i)
62 v.push_back(mp(len[i], i));
63 sort(v.begin(), v.end());
64 for (int i = 2; i < v.size(); ++i)
65 if (v[i-2].first + v[i-1].first > v[i].first)
66 {
67 printf("%lld %lld %lld\n", v[i-2].second, v[i-1].second, v[i].second);
68 return;
69 }
70 puts("-1");
71 }
72 const ll inf = 1000000000000000000;
73 int get_maxl()
74 {
75 ll a = 1, b = 1;
76 int cnt = 2;
77 while (b <= inf)
78 {
79 a += b;
80 swap(a, b);
81 ++cnt;
82 }
83 return cnt;
84 }
85 int main()
86 {
87 cerr << "maxl = " << (maxl = get_maxl()) << endl;
88 scanf("%d %d", &n, &q);
89 for (int i = 1; i <= n; ++i)
90 scanf("%lld", len+i);
91 for (int i = 1; i <= q; ++i)
92 {
93 int l, r;
94 scanf("%d %d", &l, &r);
95 solve(l, r);
96 }
97
98 }
O(n*len*log(len))预处理，O(1) 询问。
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27
28 /*
29 by skydog
30 */
31 #include <iostream>
32 #include <cstdio>
33 #include <vector>
34 #include <utility>
35 #include <algorithm>
36 #include <cmath>
37 #include <cstring>
38 #include <map>
39 #include <set>
40 #include <stack>
41 #include <queue>
42 #include <deque>
43 #include <cassert>
44 #include <list>
45 using namespace std;
46 typedef long long ll;
47 typedef pair<int, int> ii;
48 typedef pair<ll, ll> l4;
49
50 #define mp make_pair
51 #define pb push_back
52 #define db(x) cerr << #x << " = " << x << endl
53
54 const int maxn = 3e5+1;
55 ll len[maxn];
56 int ans[maxn];
57 int ret[maxn][3];
58 int maxl = 88;
59 int n, q;
60
61 const ll inf = 1000000000000000000;
62 int get_maxl()
63 {
64 ll a = 1, b = 1;
65 int cnt = 2;
66 while (b <= inf)
67 {
68 a += b;
69 swap(a, b);
70 ++cnt;
71 }
72 return cnt;
73 }
74 multiset<l4> st;
75 bool check(int from, int to, multiset<l4>::iterator a, multiset<l4>::iterator b, multiset<l4>::iterator c)
76 {
77 if (a->first + b->first > c->first)
78 {
79 ans[from] = to;
80 ret[from][0] = a->second, ret[from][1] = b->second, ret[from][2] = c->second;
81 return true;
82 }
83 return false;
84 }
85 void solve(int i)
86 {
87 st.clear();
88 int lim = min(n, i+maxl-1);
89 ans[i] = n+1;
90 for (int j = i; j <= lim; ++j)
91 {
92 st.insert(mp(len[j], j));
93 multiset<l4>::iterator it = st.find(mp(len[j], j)), it1, it2, it_1, it_2;
94 it1 = it2 = it_1 = it_2 = st.end();
95 if (it != st.end())
96 {
97 it1 = it;
98 ++it1;
99 if (it1 != st.end())
100 {
101 it2 = it1;
102 ++it2;
103 }
104 }
105 if (it != st.begin())
106 {
107 it_1 = it;
108 --it_1;
109 if (it_1 != st.begin())
110 {
111 it_2 = it_1;
112 --it_2;
113 }
114 }
115 if (it_1 != st.end())
116 {
117 if (it_2 != st.end() && check(i, j, it_2, it_1, it))
118 return;
119 if (it1 != st.end() && check(i, j, it_1, it, it1))
120 return;
121 }
122 if (it1 != st.end() && it2 != st.end() && check(i, j, it, it1, it2))
123 return;
124 }
125 }
126 int main()
127 {
128 cerr << "maxl = " << (maxl = get_maxl()) << endl;
129 scanf("%d %d", &n, &q);
130 for (int i = 1; i <= n; ++i)
131 scanf("%lld", len+i);
132 int nxt = 1;
133 for (int i = 1; i <= n; ++i)
134 solve(i);
135 for (int i = 1; i <= q; ++i)
136 {
137 int l, r;
138 scanf("%d %d", &l, &r);
139 if (r >= ans[l])
140 {
141 printf("%d %d %d\n", ret[l][0], ret[l][1], ret[l][2]);
142 }
143 else
144 {
145 puts("-1");
146 }
147 }
148
149 }
E.
题意：给你两个string的序列A[1-n], B[1-k]，1 <= n, k <= 1e5，每个字符串长度都不超过10，B中的字符串没有重复。问有多少个(l, r)，1 <= l <= r <= n满足B中的每一个元素都在A[l-r]中出现过，即set{A[l-r]} 包含 set{B[1-k]}。
观察：其实就是一个滑动窗口问题
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27 const int maxn = 1e5+1;
28 string s[maxn];
29 int idx[maxn];
30 map<string, int> id;
31 int cnt[maxn] = {0}, tot = 0;
32 int n, k;
33 int main()
34 {
35 ios::sync_with_stdio(false);
36 cin.tie(0);
37 cin >> n >> k;
38 for (int i = 1; i <= n; ++i)
39 cin >> s[i];
40 for (int i = 1; i <= k; ++i)
41 {
42 string str;
43 cin >> str;
44 id[str] = i;
45 }
46 for (int i = 1; i <= n; ++i)
47 idx[i] = id[s[i]];
48 ll ans = 0, nxt = 1;
49 cnt[0] = 1;
50 for (int i = 1; i <= n; ++i)
51 {
52 while (nxt <= n && tot < k)
53 {
54 if (cnt[idx[nxt]] == 0)
55 ++tot;
56 ++cnt[idx[nxt]];
57 ++nxt;
58 }
59 if (tot < k)
60 break;
61 ans += n-nxt+2;
62 --cnt[idx[i]];
63 if (cnt[idx[i]] == 0)
64 --tot;
65 }
66 printf("%lld\n", ans);
67
68 }
F.
题意：给你一颗大小不超过2e5的树，让你在树上选两个不同的点a和b。树上两点间的距离定义为两点间最短路径上的边数，树上一点的不稳定性定义为这个点到a或b距离的最小值，整棵树的不稳定性定义为，树上任意一点不稳定性的最大值。让你选出a，b使得树的不稳定性最小。
观察：如果选一个点的话会选在哪呢？直径中点。那两个点是不是也会和直径有关？令直径上点的个数为len，那么最小的不稳定性不会超过len/2，因为你只要在直径的中点上放一个点（如果直径中点在一条边上，就在这条边两个端点上均放一个点），那么其他点到这个点（或两个点）的最小距离一定不超过len/2。那么其实可以考虑最小的不稳定性mini，mini <= len/2。如果答案是mini，那么我们只需要选离直径上距直径两端点距离为mini的点就好了（如果有重复，选另外的点）。所以可以考虑二分答案。
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27 const int maxn = 2e5+1;
28 vector<int> g[maxn];
29 int n, pre[maxn];
30 int bfs(int cur)
31 {
32 int ret = -1;
33 memset(pre, -1, sizeof(pre));
34 pre[cur] = 0;
35 queue<int> q;
36 q.push(cur);
37 while (!q.empty())
38 {
39 int cur = q.front();
40 q.pop();
41 ret = cur;
42 for (auto nxt : g[cur])
43 if (pre[nxt] == -1)
44 {
45 pre[nxt] = cur;
46 q.push(nxt);
47 }
48 }
49 assert(ret != -1);
50 return ret;
51 }
52 vector<int> diameter;
53 int dis[maxn];
54 bool valid(int mid)
55 {
56 memset(dis, -1, sizeof(dis));
57 queue<int> q;
58 int a = diameter[mid], b = diameter[diameter.size()-1-mid];
59 dis[a] = dis[b] = 0;
60 q.push(a);
61 q.push(b);
62 while (!q.empty())
63 {
64 int cur = q.front();
65 q.pop();
66 if (dis[cur] > mid)
67 return false;
68 int nxtd = dis[cur]+1;
69 for (auto nxt : g[cur])
70 if (dis[nxt] == -1)
71 {
72 dis[nxt] = nxtd;
73 q.push(nxt);
74 }
75 }
76 return true;
77 }
78 int main()
79 {
80 scanf("%d", &n);
81 for (int i = 1; i <= n-1; ++i)
82 {
83 int u, v;
84 scanf("%d %d", &u, &v);
85 g[u].pb(v);
86 g[v].pb(u);
87 }
88 int lr = bfs(1);
89 int rr = bfs(lr);
90 //get diameter
91 {
92 int cur = rr;
93 while (cur)
94 {
95 diameter.pb(cur);
96 cur = pre[cur];
97 }
98 assert(diameter.back() == lr);
99 }
100 int half_len = diameter.size()/2;
101
102 int l = 0, r = half_len, mid, ans=0;
103 while (l <= r)
104 {
105 mid = (l+r)>>1;
106 if (valid(mid))
107 {
108 ans = mid;
109 r = mid-1;
110 }
111 else
112 l = mid+1;
113 }
114 int a = diameter[ans], b = diameter[diameter.size()-ans-1];
115 if (a == b)
116 for (int i = 1; i <= n; ++i)
117 if (i != a)
118 {
119 b = i;
120 break;
121 }
122 printf("%d %d\n", a, b);
123 }
原文地址：https://www.cnblogs.com/skyette/p/8452439.html
时间： 2024-10-13 18:46:14