思路分析
对一次区间又两种操作,要么全部1操作,要么选一个最小值进行2操作,分治解决
AC代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
bool IOerror=0;
inline char nc() {
static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
if(p1==pend) {
p1=buf;
pend=buf+fread(buf,1,BUF_SIZE,stdin);
if(pend==p1) {
IOerror=1;
return -1;
}
}
return *p1++;
}
inline bool blank(char ch) {
return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
}
template<class T> inline bool read(T &x) {
bool sign=0;
char ch=nc();
x=0;
for(; blank(ch); ch=nc());
if(IOerror)return false;
if(ch=='-')sign=1,ch=nc();
for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
if(sign)x=-x;
return true;
}
template<class T,class... U>bool read(T& h,U&... t) {
return read(h)&&read(t...);
}
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 1e4+10;
int a[N];
int solve(int l,int r){
if(l>r) return 0;
if(l==r) return a[l]==0?0:1;
int mi=INF,pos;
for(int i=l;i<=r;i++){
if(a[i]<mi){
mi = a[i];
pos = i;
}
}
for(int i=l;i<=r;i++) a[i]-=mi;
int ans = min(r-l+1,solve(l,pos-1)+solve(pos+1,r)+mi);
for(int i=l;i<=r;i++) a[i]+=mi;
return ans;
}
int main() {
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
int n;
cin>>n;
for(int i=1;i<=n;i++) cin>>a[i];
cout<<solve(1,n)<<endl;
return 0;
}