P3035 [USACO11DEC]伞头Umbrellas for Cows

此题难点在伞的价格没有单调性.
想到了背包, 但是很明显 (O(NM)) 的复杂度太高.
其实, 考虑到大的伞假如比小的伞便宜, 小的伞完全可以不考虑.
所以求一个后缀最小值就有单调性了, 然后dp就行了,,,

#include <cstdio>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAXN = 5e3 + 10;
inline int read(){
    char ch = getchar(); int x = 0;
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return x;
}

int N, M;
int a[MAXN], c[(int) 1e5 + 10];

ll f[MAXN];

int main(){
    // freopen("p3035.in", "r", stdin);
    cin>>N>>M;
    for(int i = 1; i <= N; i++) a[i] = read();
    sort(a + 1, a + N + 1);
    for(int i = 1; i <= M; i++) c[i] = read();
    for(int i = M - 1; i >= 1; i--) c[i] = min(c[i], c[i + 1]);
    memset(f, 0x7f, sizeof(f)); f[0] = 0;
    for(int i = 1; i <= N; i++) {
        for(int j = 0; j < i; j++) 
            f[i] = min(f[i], f[j] + c[a[i] - a[j + 1] + 1]);
    }
    cout<<f[N]<<endl; 
    return 0;
}