目录
题目链接题解代码
题目链接
CF868F. Yet Another Minimization Problem
题解
\(f_{i,j}=\min\limits_{k=1}^{i}\{f_{k,j-1}+w_{k,i}\}\)\(w_{l,r}\)为区间\([l,r]\)的花费,1D1D的经典形式 发现这个这是个具有决策单调性的转移 单无法快速转移,我们考虑分治 对于当前分治区间\([l,r]\) ,它的最优决策区间在\([L,R]\)之间。 对于\([l,r]\)的中点\(mid\)，我们可以暴力扫\([L−mid]\) 找到mid的最优决策点p。因为决策单调，所以\([l,mid−1]\)最优决策区间为\([L,p]\)，而\([mid+1,r]\),的最优决策区间在\([p,R]\)上 分治下去 求解区间：\(|\gets预处理\to | l\frac{\qquad\qquad\qquad\downarrow^{mid}\qquad\qquad\qquad}{}r\)
决策区间：\(L\frac{\qquad\qquad\qquad\downarrow^{p}\qquad\qquad\qquad}{}R\)
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define gc getchar()
#define pc putchar
inline int read() {
int x = 0,f = 1;
char c = gc;
while(c < '0' || c > '9' )c = gc;
while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
return x * f ;
}
void print(LL x) {
if(x >= 10) print(x / 10);
pc(x % 10 + '0');
}
int n,K;
const int maxn = 200007;
int a[maxn],b[maxn],c[maxn];
LL f[maxn],dp[maxn];
void solve(int l,int r ,int L,int R,int w) {
if(l > r) return ;
int mid = l + r >> 1,k = 0,p = std::min(mid,R);
for(int i = l;i <= mid;++ i) w += c[a[i]] ++;
for(int i = L;i <= p;++ i) {
w -= -- c[a[i]];
if(dp[mid] > f[i] + w) dp[mid] = f[i] + w,k = i;
}
for(int i = L;i <= p;++ i) w += c[a[i]] ++;
for(int i = l;i <= mid;++ i) w -= --c[a[i]];
solve(l,mid - 1,L,k,w);
for(int i = l;i <= mid;++ i) w += c[a[i]] ++;
for(int i = L;i < k;++ i) w -= -- c[a[i]];
solve(mid + 1,r,k,R,w);
for(int i = L;i < k;++ i) ++ c[a[i]];
for(int i = l;i <= mid;++ i) -- c[a[i]];
}
int main() {
n = read(),K = read();
for(int i = 1;i <= n;++ i)
f[i] = f[i - 1] + c[a[i] = read()] ++;
memset(c,0,sizeof c);
for(int i = 1;i <= K;++ i) {
memset(dp,0x3f,sizeof dp);
solve(1,n,1,n,0);
std::swap(f,dp);
}
print(dp[n]);
return 0;
}
转载于:https://www.cnblogs.com/sssy/p/9768594.html
