題目鏈接

BZOJ3594

題解

dp難題總是想不出來，，

首先要觀察到一個很重要的性質，就是每次拔高一定是拔一段後綴
因為如果單獨只拔前段的話，後面與前面的高度差距大了，不優反劣

然後很顯然可以設出\(f[i][j]\)表示前\(i\)個玉米，第\(i\)棵必須選，且共拔高了\(j\)次的最大值
由之前的性質，我們知道\(f[i][j]\)狀態中\(i\)的高度是\(h[i] + j\)
所以可以的到狀態轉移方程：
\[f[i][j] = max\{f[k][l]\} + 1 \quad [k < i] \quad [h[i] + j \ge h[k] + l]\]
可以用二維樹狀數組維護

復雜度\(O(nK + nlog^2n)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
 

#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define lbt(x) (x & -x)
using namespace std;
const int maxn = 10005,maxm = 505,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1
 
; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int h[maxn],n,K;
int s[maxn][maxm],N = 5503,M = 503;
void modify(int u,int p,int v){
    for (int i = u; i <= N; i += lbt(i))
        for (int j = p; j <= M; j += lbt(j))
            s[i][j] = max(s[i][j],v);
}
int query(int u,int p){
    int re = 0;
    for (int i = u; i; i -= lbt(i))
        for (int j = p; j; j -= lbt(j))
            re = max(re,s[i][j]);
    return re;
}
int main(){
    n = read(); K = read();
    for (int i = 1; i <= n; i++) h[i] = read();
    int ans = 0;
    for (int i = 1; i <= n; i++){
        for (int j = K; j >= 0; j--){
            int t = query(h[i] + j,j + 1) + 1;
            ans = max(ans,t);
            modify(h[i] + j,j + 1,t);
        }
    }
    printf("%d\n",ans);
    return 0;
}

BZOJ3594 [Scoi2014]方伯伯的玉米田 【樹狀數組優化dp】