Codeforces 1000 组合数可行线段倒dp 边双联通缩点求树直径
A
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } int num[15]; int anser[15]; map<string, int> mp; int main() { mp["S"] = 1, mp["M"] = 2, mp["L"] = 3; mp["XL"] = 4, mp["XXL"] = 5, mp["XXXL"] = 6; mp["XS"] = 7, mp["XXS"] = 8, mp["XXXS"] = 9; int n; int ans = 0; read(n); string now; for (int i = 1; i <= n; i++) { cin >> now; num[mp[now]]++; } for (int i = 1; i <= n; i++) { cin >> now; anser[mp[now]]++; } for (int i = 1; i <= 14; i++) { ans += abs(num[i] - anser[i]); } cout << ans / 2 << endl; return 0; }
B
分奇偶即可
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } int num[200005]; int anser = 0; int number[200005]; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n, m; cin >> n >> m; for (int i = 1; i <= n; i++) { cin >> num[i]; } n++; num[n] = m; for (int i = 1; i <= n; i++) { number[i] = number[i - 1]; if (i & 1) { int add = num[i] - num[i - 1]; number[i] += add; } } anser = number[n]; for (int i = 1; i <= n - 1; i++) { if (num[i] - 1 > num[i - 1] || num[i] + 1 < num[i + 1]) { if (i & 1) { int now = number[i] + m - num[i] + number[i] - number[n] - 1; anser = max(anser, now); } } } cout << anser << endl; return 0; }
C
离散化前缀和
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } ll number[600005]; ll ans[600005]; ll l[200005], r[200005]; ll anser[200005]; map<ll, ll> mp; map<ll, ll> mpb; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n; cin >> n; for (int i = 1; i <= n; i++) { cin >> l[i] >> r[i]; number[3 * i - 2] = l[i]; number[3 * i - 1] = r[i]; number[3 * i] = r[i] + 1; } sort(number + 1, number + 3 * n + 1); int pop = unique(number + 1, number + 3 * n + 1) - number - 1; // for (int i = 1; i <= pop; i++) // { // cout << number[i] << " "; // } // cout << endl; for (int i = 1; i <= pop; i++) { mp[number[i]] = i; mpb[i] = number[i]; } for (int i = 1; i <= n; i++) { ans[mp[l[i]]]++; ans[mp[r[i]] + 1]--; } for (int i = 1; i <= pop; i++) { ans[i] += ans[i - 1]; } ll cur = ans[1]; for (int i = 1; i <= pop - 1; i++) { cur = ans[i]; anser[cur] += mpb[i + 1] - mpb[i]; } cur = ans[pop]; anser[cur]++; for (int i = 1; i <= n; i++) { cout << anser[i] << " "; } cout << endl; return 0; }
D
倒着DP
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 998244353, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 2e5 + 5, MAXM = 2e5 + 5, N = 2e5 + 5; const int MAXQ = 100010; /*int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; }*/ inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } ll zuhe[1005][1005]; int n; ll num[1005]; ll ans[1005]; ll anser = 0; void init() { zuhe[0][0] = 1; for (int i = 1; i <= n; i++) { zuhe[i][0] = 1; for (int j = 1; j <= i; j++) { zuhe[i][j] = (zuhe[i - 1][j] + zuhe[i - 1][j - 1]) % mod; } } } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> n; ans[n + 1] = 1; init(); for (int i = 1; i <= n; i++) { cin >> num[i]; } for (int i = n; i >= 1; i--) { if (num[i] <= 0) { continue; } for (int j = i + num[i] + 1; j <= n + 1; j++) { ans[i] += zuhe[j - i - 1][num[i]] * ans[j]; ans[i] %= mod; } } for (int i = 1; i <= n; i++) { anser += 1LL * ans[i]; anser %= mod; } cout << anser << endl; return 0; }
E
边双联通分量缩点后是一颗树 再求树的直径即可
/*Huyyt*/ #include <bits/stdc++.h> #include <vector> #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <algorithm> #include <queue> #include <cstdlib> #include <stack> #include <vector> #include <set> #include <map> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int N = 300000 + 5; const int M = 600000 + 5; struct EDGE { int v, next; } edge[M * 2]; int first[N], low[N], dfn[N], belong[N], degree[N], sta[M], instack[M]; vector<int> gra[300005]; int g, cnt, top, scc; int min(int a, int b) { return a < b ? a : b; } void AddEdge(int u, int v) { edge[g].v = v; edge[g].next = first[u]; first[u] = g++; } void Tarjan(int u, int fa) { int i, v; low[u] = dfn[u] = ++cnt; sta[++top] = u; instack[u] = 1; for (i = first[u]; i != -1; i = edge[i].next) { v = edge[i].v; if (i == (fa ^ 1)) { continue; } if (!dfn[v]) { Tarjan(v, i); low[u] = min(low[u], low[v]); } else if (instack[v]) { low[u] = min(low[u], dfn[v]); } } if (dfn[u] == low[u]) { scc++; while (1) { v = sta[top--]; instack[v] = 0; belong[v] = scc; if (v == u) { break; } } } } map<pair<int, int>, int> mp; int ssssstart, eeeeeeend; int distence[300005]; int visit[300005]; void getdis(int x) { visit[x] = 1; int len = gra[x].size(); for (int i = 0; i < len; i++) { int to = gra[x][i]; if (visit[to]) { continue; } distence[to] = distence[x] + 1; getdis(to); } } int main() { ios_base::sync_with_stdio(0); cin.tie(0); memset(first, -1, sizeof(first)); int n, m; int u, v; cin >> n >> m; for (int i = 1; i <= m; i++) { cin >> u >> v; AddEdge(u, v); AddEdge(v, u); } for (int i = 1; i <= n; i++) { if (!dfn[i]) { Tarjan(i, -1); } } for (int i = 1; i <= n; i++) { for (int j = first[i]; j != -1; j = edge[j].next) { v = edge[j].v; if (belong[i] != belong[v]) { if (!mp[make_pair(belong[i], belong[v])]) { gra[belong[i]].push_back(belong[v]); gra[belong[v]].push_back(belong[i]); mp[make_pair(belong[i], belong[v])] = mp[make_pair(belong[v], belong[i])] = 1; } } } } getdis(1); int maxnn = -1; for (int i = 1; i <= scc; i++) { if (distence[i] > maxnn) { ssssstart = i; maxnn = distence[i]; } } mem(visit, 0), mem(distence, 0); getdis(ssssstart); maxnn = -1; for (int i = 1; i <= scc; i++) { if (distence[i] > maxnn) { eeeeeeend = i; maxnn = distence[i]; } } cout << distence[eeeeeeend] << endl; return 0; }
