Codeforces 1077 F2 - Pictures with Kittens (hard version)

F2 - Pictures with Kittens (hard version)

思路：

单调队列优化dp

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair
#define pli pair
#define pii pair
#define piii pair
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 5e3 + 10;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int a[N];
dequeint, LL>> dp[N]; 
LL tmp[N];
int main() {
    int n, k, x;
    scanf("%d %d %d", &n, &k, &x);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);

        dp[0].pb({0, 0});
        for (int i = 1; i <= n; i++) {
            // i-k 
            for (int l = 1; l <= x; l++) {
                while(!dp[l-1].empty() && dp[l-1].front().fi < i-k) dp[l-1].pop_front();
                if(!dp[l-1].empty()) tmp[l] = dp[l-1].front().se + a[i];
            }
            
            for (int l = 1; l <= x; l++) {
                while(!dp[l].empty() && dp[l].back().se <= tmp[l]) dp[l].pop_back();
                if(tmp[l])dp[l].push_back({i, tmp[l]});
                tmp[l] = 0;
            }
        }
    LL ans = -1; 
    while(!dp[x].empty() && dp[x].front().fi <= n-k)  dp[x].pop_front();
    if(!dp[x].empty()) ans = dp[x].front().se; 
    printf("%lld\n", ans);
    return 0;
}

 

转载于:https://www.cnblogs.com/widsom/p/9972681.html