1563: [NOI2009]诗人小G
https://lydsy.com/JudgeOnline/problem.php?id=1563
分析:
直接转移f[i]=f[j]+cost(i,j),cost(i,j)=(sum[i]-sum[j])p
然后有决策单调性,就可以二分+队列了。注意两个字符串之间还有一个空格,所以长度+1,很多字符串合起来后,总的长度还要-1,最后一个没空格。
证明?byvoid
luogu输出方案,加上后一直过不了,g,nxt数组是输出方案的部分
代码:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#define fi(s) freopen(s,"r",stdin);
#define fo(s) freopen(s,"w",stdout);
using namespace std;
typedef long long LL;
typedef long double LD; inline int read() {
int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
} const int N = ;
const LL INF = 1e18; char str[N][], pr[];
int len[N], g[N], nxt[N];
LD f[N];
LL sum[N];
int n, Len, P;
struct Node{
int p, l, r;
Node() {}
Node(int _,int __,int ___) { p = _, l = __, r = ___; }
}q[N]; LD ksm(LD x) {
LD ans = ; int b = P;
while (b) {
if (b & ) ans = ans * x;
x = x * x;
b >>= ;
}
return ans;
}
void init() {
n = read(), Len = read(), P = read();
for (int i=; i<=n; ++i) {
scanf("%s", str[i] + );
sum[i] = sum[i - ] + (len[i] = strlen(str[i] + ) + );
}
}
LD Calc(int i,int j) {
return f[j] + ksm(abs(sum[i] - sum[j] - Len - ));
}
int binary_search(int l,int r,int x,int y) {
int ans = n;
while (l <= r) {
int mid = (l + r) >> ;
if (Calc(mid, x) < Calc(mid, y)) ans = mid, r = mid - ;
else l = mid + ;
}
return ans;
}
void solve() {
int L = , R = ;
q[++R] = Node(, , n);
for (int i=; i<=n; ++i) {
while (L <= R && q[L].r < i) L ++;
int j = q[L].p;
f[i] = Calc(i, j);
g[i] = j;
if (Calc(n, q[R].p) < Calc(n, i)) continue; // 如果最后一个点从i转移不优,那么说明i没有覆盖的区间,不需要二分了
while (L <= R && Calc(q[R].l, q[R].p) > Calc(q[R].l, i)) R --;
q[R].r = binary_search(q[R].l, n, i, q[R].p) - ;
q[++R] = Node(i, q[R - ].r + , n);
}
}
void print() {
if (f[n] > INF) puts("Too hard to arrange");
else printf("%lld\n", (LL)(f[n]));
puts(pr);
}
int main() {
strcpy(pr, "--------------------");
int T = read();
while (T--) {
init();
solve();
print();
}
return ;
}