Codeforces round #267 (Div. 2)

Codeforces round #267 (Div. 2)

A: sign-in question, directly for once

B: Take an exception or a different number, and then bitcount to judge it.

C: DP, DP [I] indicates the maximum value to I, and then you can transfer the obtained and not the current position. You must first prefix and pre-process them.

D: first, use map to hash the string, and then create a graph. During the on-site operation, the search is directly memorized. However, this will not solve the ring problem and the FST will be decisive, later, I changed my posture. I first sought strong connectivity to scale down the point. After the scale-down, I got a dag, And then I needed to perform a memory-based search.

Code:

A:

#include <cstdio>int main() {int n, p, q;int ans = 0;scanf("%d", &n);while (n--) {scanf("%d%d", &p, &q);if (q - p >= 2) ans++;}printf("%d\n", ans);return 0;}

B:

#include <cstdio>#include <cstring>int n, m, k;int bitcount(int x) {int ans = 0;while (x) {ans += (x&1);x >>= 1;}return ans;}const int N = 1005;int x[N];int main() {scanf("%d%d%d", &n, &m, &k);for (int i = 0; i <= m; i++)scanf("%d", &x[i]);int ans = 0;for (int i = 0; i < m; i++) {if (bitcount(x[i]^x[m]) <= k) ans++;}printf("%d\n", ans);return 0;}

C:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 5005;const long long INF = 0x3f3f3f3f3f3f3f;long long dp[N][N];int n, m, k;long long num[N], pre[N];int main() {scanf("%d%d%d", &n, &m, &k);for (int i = 1; i <= n; i++) {scanf("%lld", &num[i]);pre[i] = pre[i - 1] + num[i];}for (int i = 0; i <= n; i++)for (int j = 0; j <= k; j++)dp[i][j] = -INF;dp[0][0] = 0;for (int i = 1; i <= n; i++) {for (int j = 0; j <= k; j++) {if (j && i >= m)dp[i][j] = max(dp[i][j], dp[i - m][j - 1] + pre[i] - pre[i - m]);dp[i][j] = max(dp[i][j], dp[i - 1][j]);}}printf("%lld\n", dp[n][k]);return 0;}

D:

#include <cstdio>#include <cstring>#include <vector>#include <stack>#include <algorithm>#include <string>#include <iostream>#include <map>using namespace std;typedef long long ll;const int N = 1000005;vector<int> g[N], scc[N];int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt;stack<int> S;typedef pair<ll, ll> pii;int n, m, hn;map<string, int> hash;string str[N];pii s[N];pii p[N];pii dp[N];int vis[N];vector<int> g2[N];void dfs_scc(int u) {pre[u] = lowlink[u] = ++dfs_clock;S.push(u);for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (!pre[v]) {dfs_scc(v);lowlink[u] = min(lowlink[u], lowlink[v]);} else if (!sccno[v])lowlink[u] = min(lowlink[u], pre[v]);}if (lowlink[u] == pre[u]) {scc_cnt++;pii tmp = s[S.top()];while (1) {int x = S.top(); S.pop();tmp = min(tmp, s[x]);sccno[x] = scc_cnt;if (x == u) break;}p[scc_cnt] = tmp;}}void find_scc(int n) {dfs_clock = scc_cnt = 0;memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));for (int i = 0; i < n; i++)if (!pre[i]) dfs_scc(i);for (int i = 0; i < n; i++) {for (int j = 0; j < g[i].size(); j++) {int u = sccno[i], v = sccno[g[i][j]];if (u == v) continue;g2[u].push_back(v);}}}int get(string& str) {for (int i = 0; i < str.length(); i++)if (str[i] >= 'A' && str[i] <= 'Z')str[i] = str[i] - 'A' + 'a';if (!hash.count(str)) {hash[str] = hn;int cnt = 0;for (int i = 0; i < str.length(); i++)if (str[i] == 'r') cnt++;s[hn].first = cnt;s[hn].second = str.length();hn++;}return hash[str];}pii dfs(int u) {if (vis[u]) return dp[u];vis[u] = 1;dp[u] = p[u];for (int i = 0; i < g2[u].size(); i++) {int v = g2[u][i];dp[u] = min(dp[u], dfs(v));}return dp[u];}int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {cin >> str[i];get(str[i]);}scanf("%d", &m);string u, v;while (m--) {cin >> u >> v;int uu = get(u);int vv = get(v);g[uu].push_back(vv);}find_scc(hn);ll ans1 = 0, ans2 = 0;for (int i = 0; i < n; i++) {int u = sccno[get(str[i])];dfs(u);ans1 += dp[u].first;ans2 += dp[u].second;}cout << ans1 << " " << ans2 << endl;return 0;}

Codeforces round #267 (Div. 2)