A
线段覆盖，求未覆盖长度
左端点排序，然后扫一遍
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define endl "/n"
5 #define mp make_pair
6 #define pb push_back
7 #define st first
8 #define nd second
9 #define pii pair<int, int>
10
11 #define sz(x) (int((x).size())
12 #define rep(i, s, t) for (int i = (int)(s); i <= (int)(t); ++ i)
13 #define inc(i, s, t) for (int i = (int)(s); i < (int)(t); ++ i)
14
15 typedef long long ll;
16 typedef long double ld;
17 typedef unsigned long long ull;
18
19 #ifdef LOCAL
20 ostream &operator<<(ostream &out, string str) {
21 for (char c : str)
22 out << c;
23 return out;
24 }
25 template <class L, class R> ostream &operator<<(ostream &out, pair<L, R> p) { return out << "(" << p.st << ", " << p.nd << ")"; }
26 template <class L, class R, class S> ostream &operator<<(ostream &out, tuple<L, R, S> p) {
27 auto &[a, b, c] = p;
28 return out << "(" << a << ", " << b << ", " << c << ")";
29 }
30 template <class T> auto operator<<(ostream &out, T a) -> decltype(a.begin(), out) {
31 out << '{';
32 for (auto it = a.begin(); it != a.end(); it = next(it))
33 out << (it != a.begin() ? ", " : "") << *it;
34 return out << '}';
35 }
36 void dump() { cerr << "/n"; }
37 template <class T, class... Ts> void dump(T a, Ts... x) {
38 cerr << a << ", ";
39 dump(x...);
40 }
41 #define debug(...) cerr << "[" #__VA_ARGS__ "]: ", dump(__VA_ARGS__)
42 #else
43 #define debug(...) ;
44 #endif
45
46 const int N = 2e5 + 10;
47 #define int long long
48 struct Node {
49 int x, r;
50 bool operator < (const Node& rhs) const {
51 if (x == rhs.x) return x + r < rhs.x + rhs.r;
52 return x < rhs.x;
53 }
54 } a[N];
55
56 int32_t main() {
57 int n; cin >> n;
58 rep(i, 1, n) cin >> a[i].x >> a[i].r;
59 sort(a + 1, a + n + 1);
60 ll ans = 0, id = 1;
61 rep(i, 2, n) {
62 debug(id);
63 debug(a[id].x, a[id].r);
64 if (a[id].x + a[id].r >= a[i].x - a[i].r) {
65 if (a[i].x + a[i].r > a[id].x + a[id].r)
66 id = i;
67 } else {
68 ans += a[i].x - a[i].r - (a[id].x + a[id].r);
69 id = i;
70 }
71 }
72 cout << ans << endl;
73 }
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D
结论题，考虑如何能得到最大圆弧
那么发现当旋转的线段所在的直线过圆心时，圆弧上的曲率最大
相同长度上的圆弧，曲率越大，圆弧越长
所以直接算出直线过原点的情况即可
1 #include <bits/stdc++.h>
2 using namespace std;
3 double r;
4 double x,y,d;
5 int main(){
6 int T;scanf("%d",&T);
7 while(T--){
8 cin>>r>>x>>y>>d;
9 double D=sqrt(x*x+y*y);
10 double a=asin((d+D)/r),b=asin((d-D)/r);
11 printf("%.12lf/n",a*r+b*r);
12 }
13 return 0;
14 }
View Code
G
签到题
题意就是给一堆区间把所有相交的区间合并，
找出合并后，所有区间后的空隙
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define LOCAL
4
5 #define endl "/n"
6 #define mp make_pair
7 #define pb push_back
8 #define st first
9 #define nd second
10 #define pii pair<int, int>
11
12 #define sz(x) (int((x).size())
13 #define rep(i, s, t) for (int i = int(s); i <= int(t); ++ i)
14 #define inc(i, s, t) for (int i = int(s); i < int(t); ++ i)
15
16 typedef long long ll;
17 typedef long double ld;
18 typedef unsigned long long ull;
19
20 #ifdef LOCAL
21 ostream &operator<<(ostream &out, string str) {
22 for (char c : str)
23 out << c;
24 return out;
25 }
26 template <class L, class R> ostream &operator<<(ostream &out, pair<L, R> p) { return out << "(" << p.st << ", " << p.nd << ")"; }
27 template <class L, class R, class S> ostream &operator<<(ostream &out, tuple<L, R, S> p) {
28 auto &[a, b, c] = p;
29 return out << "(" << a << ", " << b << ", " << c << ")";
30 }
31 template <class T> auto operator<<(ostream &out, T a) -> decltype(a.begin(), out) {
32 out << '{';
33 for (auto it = a.begin(); it != a.end(); it = next(it))
34 out << (it != a.begin() ? ", " : "") << *it;
35 return out << '}';
36 }
37 void dump() { cerr << "/n"; }
38 template <class T, class... Ts> void dump(T a, Ts... x) {
39 cerr << a << ", ";
40 dump(x...);
41 }
42 #define debug(...) cerr << "[" #__VA_ARGS__ "]: ", dump(__VA_ARGS__)
43 #else
44 #define debug(...) ;
45 #endif
46
47 int32_t main() {
48 string s;
49 cin >> s;
50 int n = s.length();
51 if (n == 1) cout << s;
52 else {
53 bool flag = true;
54 inc(i, 0, n - 1) if (s[i] != '9') flag = false;
55 if (flag) cout << s;
56 else {
57 inc(i, 0, n - 1) printf("9");
58 }
59 }
60 }
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I
定义dp[i][j]为抽取i次牌后剩余j张单牌的概率
转移方程写为
1 dp[0][cnt]=1; 2 for(int i=1;i<=121;++i) 3 for(int j=1;j<=13;j+=2) 4 dp[i][j]=dp[i-1][j]*(124-i-3*j)*inv(124-i)+dp[i-1][j+2]*(3*j+6)*inv(124-i);
最终答案的求和
1 int ans=0; 2 for(int i=0;i<=121;++i) 3 ans=ans+i*dp[i-1][1]*3*inv(124-i);
J
启发式合并
题意是求出一个图中染色的最大点数
考虑通过边的关系将x和y进行合并，边是单向的，
有结论如果选择x能将y染色，那么选择x一定是更优的
根据这一特点建树，求出森林中最大size的树即可
1 set<int>to[N], from[N];
2 int fa[N], sz[N];
3 int findf(int x) {
4 if (fa[x] == x)return x;
5 return fa[x] = findf(fa[x]);
6 }
7 void _Merge(int x, int y) {
8 x = findf(x), y = findf(y);
9 if (x == y)return;
10 if (to[x].size() < to[y].size())
11 swap(x, y);
12 fa[y] = x;
13 sz[x] += sz[y];
14 vector<pair<int, int> >mg;
15 for (auto t : to[y]) {
16 to[x].insert(t);
17 from[t].erase(y);
18 from[t].insert(x);
19 if (from[t].size() == 1)
20 mg.push_back(make_pair(t, x));
21 }
22 for (auto [x, y] : mg)
23 _Merge(x, y);
24 }
25
26 int main() {
27 int T = fast_IO::read();
28 for (int ti = 1; ti <= T; ++ti) {
29 int n = fast_IO::read();
30 for (int i = 1; i <= n; ++i)fa[i] = i, sz[i] = 1;
31 for (int i = 1; i <= n; ++i) {
32 int k = fast_IO::read();
33 for (int j = 1; j <= k; ++j) {
34 int y = fast_IO::read();
35 to[y].insert(i);
36 from[i].insert(y);
37 }
38 }
39 for (int i = 1; i <= n; ++i)
40 if (from[i].size() == 1)
41 _Merge(*from[i].begin(), i);
42 int ans = 0;
43 for (int i = 1; i <= n; ++i)
44 ans = max(ans, sz[i]);
45 printf("Case #%d: %d/n", ti, ans);
46 for (int i = 1; i <= n; ++i)
47 to[i].clear(), from[i].clear();
48 }
49 return 0;
50 }
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原创文章，作者：bd101bd101，如若转载，请注明出处：https://blog.ytso.com/275618.html