Codeforces Round #286 (Div. 1) C, D,

Codeforces Round #286 (Div. 1) C, D,

C: The number of steps in the question seems to be many. In fact, the maximum number of steps increases by about 250, because the number of moving steps is 1 + 2 + 3 + .. n, so there will be only sqrt (n) steps, so dp [I] [j] indicates that the value of step j is increased at the I position, and then transferred.

D: in fact, for a Unicom block, a maximum of n sides and a minimum of n-1 are required. The condition for judging is whether the UNICOM block has a ring, use topological sorting to determine

Code:

C:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 30005;int n, d, cnt[N], dp[N][500];int dfs(int u, int cha) {    if (u > 30000) return 0;    if (dp[u][cha] != -1) return dp[u][cha];    int tmp = d + cha - 250;    dp[u][cha] = cnt[u];    int ans = 0;    if (tmp > 1)ans = max(ans, dfs(u + tmp - 1, cha - 1));    ans = max(ans, dfs(u + tmp, cha));    ans = max(ans, dfs(u + tmp + 1, cha + 1));    dp[u][cha] += ans;    return dp[u][cha];}int main() {    scanf("%d%d", &n, &d);    int tmp;    for (int i = 0; i < n; i++) {scanf("%d", &tmp);cnt[tmp]++;    }    memset(dp, -1, sizeof(dp));    printf("%d\n", dfs(d, 250));    return 0;}

D:

#include <cstdio>#include <cstring>#include <vector>#include <queue>using namespace std;const int N = 100005;int n, m, du[N], vis[N], have[N], hn;vector<int> g[N], g2[N];void dfs(int u) {    have[hn++] = u;    vis[u] = 1;    for (int i = 0; i < g[u].size(); i++) {        int v = g[u][i];        if (vis[v]) continue;        dfs(v);     }}bool find() {    queue<int> Q;    for (int i = 0; i < hn; i++)        if (!du[have[i]]) Q.push(have[i]);    while (!Q.empty()) {        int u = Q.front();        Q.pop();        int sz = g2[u].size();        for (int i = 0; i < sz; i++) {            int v = g2[u][i];            du[v]--;            if (!du[v]) Q.push(v);        }    }    for (int i = 0; i < hn; i++)        if (du[have[i]]) return true;    return false;}int main() {    scanf("%d%d", &n, &m);    int u, v;    while (m--) {        scanf("%d%d", &u, &v);        du[v]++;        g[u].push_back(v);        g[v].push_back(u);        g2[u].push_back(v);    }    int ans = n;    for (int i = 1; i <= n; i++) {        if (!vis[i]) {            hn = 0;            dfs(i);            if (!find()) ans--;        }    }    printf("%d\n", ans);    return 0;}