I see the solution to a problem big brother, why the first \ (1 \) sub-tasks with \ (DP \) count it? You can use greedy ah. . .
Closer to home.
For the first sub-task, direct greedy, how many can take much to take, and then take after take full cleared.
For the second sub-task, considering \ (DP \) .
Set \ (f_i \) before the representation \ (i \) a minimum value of a risk of believers, there are:
\(f_i=\min\{f_j+danger_{i,j}\}\) \((danger_{i,j}<=k)\)
Wherein \ (danger_ {i, j} \) represents \ (i, j \) dangerous values between.
The question is: \ (danger_ {i, J} \) how find it?
\ (1 \) violence to solve the equation when running.
Time complexity: \ (O (n-^. 3) \) , obviously \ (T \) .
\ (2 \) pretreatment.
Order \ (CNT \) as a counter, when the discovery of new religious \ (CNT ++ \) , and mark this religion, and each section corresponding to \ (danger_ {i, j} \) is the \ (CNT \)
Time complexity: \ (O (^ n-2) \) , it can be over.
As for the border, apparently \ (f_0 = 0, f_1 = 1 \)
The goal is naturally \ (f_n \)
\ (The AC \) \ (Code \) (code \ (sum_ {i, j} \) is the \ (danger_ {I, J} \) )
#include <iostream>
#include <cstring>
using namespace std;
int n, m, k, a[10010], cnt, ans, sum[1010][1010], f[1010];
bool vis[10010];
int main() {
cin >> n >> m >> k;
for (int i=1; i<=n; i++)
cin >> a[i];
for (int i=1; i<=n; i++) {
cnt = 0;
memset(vis, false, sizeof(vis));// 一定要清空!
for (int j=i; j<=n; j++) {
if (!vis[a[j]]) {
vis[a[j]] = true;
cnt++;
}
sum[i][j] = cnt;
}
}
cnt = 0;
memset(vis, false, sizeof(vis));//注意，出来时也要清空！我被坑了好长时间TAT
for (int i=1; i<=n; i++) {
if (!vis[a[i]]) {
vis[a[i]] = true;
cnt++;
}
if (cnt > k) {
ans++;
cnt = 1;
memset(vis, false, sizeof(vis));//记得清空！
vis[a[i]] = true;
}
if (i == n)
ans++;
}
cout << ans << endl;
f[0] = 0;
f[1] = 1;//初始化莫忘掉
for (int i=2; i<=n; i++) {
f[i] = 1e9;//因为取最小值，所以初值要赋大
for (int j=1; j<=i; j++)
if (sum[j][i] <= k) //而且危险值不能超过k
f[i] = min(f[i], f[j-1]+sum[j][i]);
}
cout << f[n] << endl;
return 0; //完结撒花！
}