C:题目中步数看似很多,其实最多就增长250步左右,由于移动的步数为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,然后转移便可
D:这题其实对1个联通块,最多只需要n条边,最少要n - 1条,那末判断的条件,就是这个联通块是不是有环,利用拓扑排序去判便可
代码:
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] != ⑴) 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, ⑴, sizeof(dp));
printf("%d
", 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
", ans);
return 0;
}