[NOI2009]诗人小G 四边形优化DP

题目传送门

f[i] = min(f[j] + val(i,j);

其中val(i,j) 满足 四边形dp策略。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
char s[N][55];
int sum[N];
int ans[N];
long double f[N];
struct Node{
    int j, l, r;
}q[N];
int n, l, p;
LL maxxx = 1e18;
long double val(int i, int j){
    if(i > j) swap(i,j);
    long double tt = abs(sum[j]-sum[i]+j-i-1-l);
    long double zz = 1;
    for(int i = 1; i <= p; ++i){
        zz *= tt;
    }
    zz += f[i];
    return zz;
}
void solve(int n){
    if(n == 0) return ;
    solve(ans[n]);
    for(int i = ans[n]+1; i <= n; ++i){
        printf("%s%c",s[i]," 
"[i==n]);
    }
}
int main(){
//    cout << (702871197447575ll<1e18) << endl;
    int T;
    scanf("%d", &T);
    for(int _ = 1; _ <= T; ++_){
        scanf("%d%d%d", &n, &l, &p);
        for(int i = 1; i <= n; ++i){
            scanf("%s", s[i]);
            sum[i] = sum[i-1] + strlen(s[i]);
        }
        int L = 1, R = 1;
        q[L] = {0,1,n};
        int midl;
        for(int i = 1; i <= n; ++i){
            q[L].l = i;
            while(L <= R && q[L].r < i) ++L;
            ans[i] = q[L].j;
            f[i] =  val(q[L].j,i);
            midl = n+1;
            while(L <= R && val(q[R].j, q[R].l) >= val(i, q[R].l)){
                R--;
            }
            if(L <= R){
                int ll = q[R].l, rr = q[R].r;
                while(ll <= rr){
                    int mid = (ll+rr) >> 1;
                    if(val(q[R].j,mid) < val(i, mid)) ll = mid+1;
                    else rr = mid - 1;
                }
                if(ll <= n){
                    q[R].r = ll-1;
                    q[++R] = {i,ll,n};
                }
            }
            else {
                q[++R] = {i,i+1,n};
            }
        }
        if(f[n] > maxxx)
            puts("Too hard to arrange");
        else{
            LL fans = f[n];
            printf("%lld
", fans);
            solve(n);
        }
        puts("--------------------");
    }
    return 0;
}