单调队列优化 dp
The only difference between easy and hard versions is the constraints.
Vova likes pictures with kittens. The news feed in the social network he uses can be represented as an array of n
consecutive pictures (with kittens, of course). Vova likes all these pictures, but some are more beautiful than the others: the i-th picture has beauty ai
.
Vova wants to repost exactly x
pictures in such a way that:
- each segment of the news feed of at least k
- consecutive pictures has at least one picture reposted by Vova;
- the sum of beauty values of reposted pictures is maximum possible.
For example, if k=1
then Vova has to repost all the pictures in the news feed. If k=2
then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.
Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.
The first line of the input contains three integers n,k
and x (1≤k,x≤n≤5000
) — the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.
The second line of the input contains n
integers a1,a2,…,an (1≤ai≤109), where ai is the beauty of the i
题意 ： 给你 n 个数字，要求从中选出 x 个数字，但任意连续的长度为 k 的区间中必须至少选择一个元素，询问所选择元素的最大的和是多少？
思路分析 ：
定义 dp[i][j] 表示前 i 个树中选择 j 个数的最大得分
代码示例 ：
n = 200
#define ll long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
ll n, k, x;
ll a[205];
ll dp[205][205];
void solve() {
memset(dp, -1*inf, sizeof(dp));
//printf("%lld ++++\n", dp[0][0]);
dp[0][0] = 0;
for(ll i = 1; i <= n; i++){
for(ll j = max(0ll, i-k); j <= i-1; j++){
for(ll f = 1; f <= x; f++){
dp[i][f] = max(dp[i][f], dp[j][f-1]+a[i]);
//prllf("++++ %d %d %d %d\n", i, j, f, dp[i][f]);
}
}
}
ll ans = -1;
for(ll i = n; i > n-k; i--) ans = max(ans, dp[i][x]);
printf("%lld\n", ans);
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
cin >> n >> k >> x;
for(ll i = 1; i <= n; i++){
scanf("%lld", &a[i]);
}
solve();
return 0;
}
n = 5000
#define ll long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
typedef pair<ll, ll> P; // pos val
#define fi first
#define se second
ll n, k, x;
ll a[5005];
deque<P>que[5005];
ll dp[5005][5005];
void solve() {
memset(dp, -1*inf, sizeof(dp));
dp[0][0] = 0;
que[0].push_back(P(0, 0));
ll ans = -1;
for(ll i = 1; i <= n; i++){
ll pos = max(i-k, 0ll);
for(ll j = 1; j <= x; j++){
while(!que[j-1].empty() && que[j-1].front().fi < pos){
que[j-1].pop_front();
}
}
for(ll j = 1; j <= x; j++){
if (!que[j-1].empty()) {
ll p = que[j-1].front().fi;
ll val = que[j-1].front().se;
dp[i][j] = max(dp[i][j], dp[p][j-1]+a[i]);
//printf("^^^^^^^^^^^ %lld %lld %lld ++++ %lld %lld %lld\n", i, j, dp[i][j], p, j-1, dp[p][j-1]);
}
}
for(ll j = 1; j <= x; j++){
while(!que[j].empty() && dp[i][j] >= que[j].back().se) que[j].pop_back();
if (dp[i][j] > 0) que[j].push_back(P(i, dp[i][j]));
if (i > n-k) ans = max(ans, dp[i][j]);
}
//for(ll j = 1; j <= x; j++) {
//printf("++++ %lld %lld %lld\n", i, j, dp[i][j]);
//}
}
printf("%lld\n", ans);
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
cin >> n >> k >> x;
for(ll i = 1; i <= n; i++){
scanf("%lld", &a[i]);
}
solve();
return 0;
}
