Codeforces Round #286 (Div. 1) C、D

C：题目中步数看似很多，其实最多就增长250步左右，因为移动的步数为1 + 2 + 3 + .. n，所以大概只会有sqrt（n）步，所以dp[i][j]表示在i位置，增长为j步的值，然后转移即可

D：这题其实对于一个联通块，最多只需要n条边，最少要n - 1条，那么判断的条件，就是这个联通块是否有环，利用拓扑排序去判即可

代码：

C：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 30005;

int n, d, cnt
, dp
[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
, vis
, have
, hn;
vector<int> g
, g2
;

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;
}