Description
https://www.luogu.org/problemnew/show/P1912
Solution
設 dpi d p i 表示到第 i i 句詩的最小不協排程,那麼顯然有:
dpi=minj=1i−1dpj+|l−sumi+sumj−i+j+1|pdpi=minj=1i−1dpj+|l−sumi+sumj−i+j+1|p
這是一個具有決策單調性的1D/1D動态規劃,我們每處理完一個 dpi d p i ,就二分其決策區間,舉個栗子:
一開始序列的決策點為:
0000000000
。
得到 dp1 d p 1 後,決策點變成了:
0011111111
。
得到 dp2 d p 2 後變成了醬紫:
0011122222
。
得到 dp3 d p 3 後,有可能是:
0011122233
,也有可能是
0013333333
。
我們二分得到決策區間的分界線、用棧來維護即可。
Code
/************************************************
* Au: Hany01
* Date: Aug 5th, 2018
* Prob: BZOJ1563 詩人小G
* Email: [email protected]
* Inst: Yali High School
************************************************/
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia
template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, : ; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, : ; }
inline int read() {
static int _, __; static char c_;
for (_ = , __ = , c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << ) + (_ << ) + (c_ ^ );
return _ * __;
}
const int maxn = + ;
LD dp[maxn];
int n, l, p, sum[maxn], pre[maxn], top, nxt[maxn];
PII stk[maxn];
char str[maxn][];
inline LD Pow(LD a, int b) {
static LD Ans;
for (Ans = ; b; b >>= , a *= a) if (b & ) Ans *= a;
return Ans;
}
inline LD calc(int i, int j) { return dp[j] + Pow((LD)abs(l - sum[i] + sum[j] - i + j + ), p); }
int main()
{
#ifdef hany01
freopen("bzoj1563.in", "r", stdin), freopen("bzoj1563.out", "w", stdout);
#endif
for (static int Cas = read(), pos; Cas --; ) {
n = read(), l = read(), p = read();
For(i, , n) scanf("%s", str[i]), sum[i] = sum[i - ] + strlen(str[i]), dp[i] = ;
stk[top = ] = mp(, );
For(i, , n) {
dp[i] = calc(i, pos = stk[upper_bound(stk + , stk + + top, mp(i, INF)) - stk - ].y);
pre[i] = pos;
if (calc(n, i) <= calc(n, stk[top].y)) {
stk[++ top] = mp(n, i);
for (register int l, r, mid; ; ) {
l = max(stk[top - ].x, i + ), r = n;
while (l < r) {
mid = (l + r) >> ;
if (calc(mid, stk[top - ].y) >= calc(mid, i)) r = mid; else l = mid + ;
}
stk[top].x = l;
if (stk[top].x > stk[top - ].x) break;
stk[top - ] = stk[top], -- top;
}
}
}
if (dp[n] > + ) puts("Too hard to arrange");
else {
printf("%lld\n", (LL)dp[n]);
for (int t = n; t; t = pre[t]) nxt[pre[t]] = t;
for (int t = ; t < n; t = nxt[t])
For(i, t + , nxt[t]) printf("%s%c", str[i], i == nxt[t] ? '\n' : ' ');
}
puts("--------------------");
}
return ;
}