标签：include || mat oid 一段 复杂度 splay eof getc
dp难题总是想不出来，，
首先要观察到一个很重要的性质，就是每次拔高一定是拔一段后缀
因为如果单独只拔前段的话，后面与前面的高度差距大了，不优反劣
然后很显然可以设出\(f[i][j]\)表示前\(i\)个玉米，第\(i\)棵必须选，且共拔高了\(j\)次的最大值
由之前的性质，我们知道\(f[i][j]\)状态中\(i\)的高度是\(h[i] + j\)
所以可以的到状态转移方程：
\[f[i][j] = max\{f[k][l]\} + 1 \quad [k < i] \quad [h[i] + j \ge h[k] + l]\]
可以用二维树状数组维护
复杂度\(O(nK + nlog^2n)\)
#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 【树状数组优化dp】
标签：include || mat oid 一段 复杂度 splay eof getc
原文地址：https://www.cnblogs.com/Mychael/p/9038058.html
