题意

题目链接

Sol

很显然的一个dp方程

\(f_i = min(f_j + (sum_i - sum_j - 1 - L)^P)\)

其中\(sum_i = \sum_{j = 1}^i len_j + 1\)

这个东西显然是有决策单调性的。

单调队列优化一下

我好像已经做过三个这种类型的题了，而且转移的时候\(w\)中总是带个幂函数。。interesting

#include<bits/stdc++.h>

#define chmax(a, b) (a = (a > b ? a : b))

#define chmin(a, b) (a = (a < b ? a : b))

#define LL long long

#define LDB long double

//#define int long long

using namespace std;

const int MAXN = 1e5 + 10;

inline int read() {

    int x = 0, f = 1; char c = getchar();

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int T, N, L, P, sum[MAXN], q[MAXN], c[MAXN], pre[MAXN];//c???ߵ?λ?

char str[MAXN][35];

LDB f[MAXN];

LDB fastpow(LDB a, int p) {

	LDB base = 1;

	while(p) {

		if(p & 1) base = base * a;

		a = a * a; p >>= 1;

	}

	return base;

}

LDB calc(int j, int i) {

	return f[j] + fastpow(abs(sum[i] - sum[j] - L), P);

}

int lower(int x, int y) {//???x????????

	int l = x, r = N + 1, ans = 0;

	while(l <= r) {

		int mid = l + r >> 1;

		if(calc(x, mid) >= calc(y, mid)) r = mid - 1;

		else l = mid + 1;

	}

	return l;

}

void solve() {

	N = read(); L = read() + 1; P = read();

	for(int i = 1; i <= N; i++) {

		scanf("%s", str[i] + 1);

		sum[i] = sum[i - 1] + strlen(str[i] + 1) + 1;

	}

	memset(q, 0, sizeof(q));

	for(int i = 1, h = 2, t = 2; i <= N; i++) {

		while(h < t && c[h] <= i) h++;

		f[i] = calc(q[h], i); pre[i] = q[h];

		while(h < t && c[t - 1] >= lower(q[t], i)) t--;

		c[t] = lower(q[t], i); q[++t] = i;

	}

	if(f[N] > 1e18) {puts("Too hard to arrange\n--------------------"); return;}

	printf("%.0Lf\n", f[N]);

    puts("--------------------");

}

main() {

	for(T = read(); T; T--) solve();

}