#10181. 「一本通 5.5 练习 2」绿色通道

${\color{cyan}{>>Question}}$

二分+$dp$(与跳房子类似)

二分答案(窗口大小),$dp$确定是否可行

$f[i]$表示到$i$($i$选)花的最少时间

$$f[i] = min_{j\in[i-d-1,i-1]}\left \{ f[j] \right \}+a[i]$$

$d$为二分的答案

注意统计是要$f[n+1]>t$才$return\;0$

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std; 

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//-------------------------------------------------------

const int N = 5e4+7;
int n,t;
int a[N],f[N];
int q[N],hd,tl,lst;

bool chk(int d) {
    memset(f,0x3f,sizeof(f)); f[0] = 0;
    hd = 1,tl = 0,lst = 0; int i;
    for(i = 1;i <= n+1; ++i) {
        while(lst < i-d-1) ++lst;
        for(;lst < i; ++lst) {
            while(hd <= tl && f[q[tl]] > f[lst]) --tl;//debug --tl -> --q[tl]
            q[++tl] = lst;
        }
        while(hd <= tl && q[hd] < i-d-1) ++hd;
        f[i] = f[q[hd]] + a[i];
        //if(f[i] > t) return 0;//debug
    }
    if(f[n+1] > t) return 0;//debug 
    return 1;
}

void bs_ans() {
    int l = 0,r = n,ans;
    while(l <= r) {
        int mid = (l+r)>>1;
        if(chk(mid)) ans = mid,r = mid-1;
        else l = mid+1;
    }
    out(ans);
}

int main() {
    in(n); in(t); int i;
    for(i = 1;i <= n;++i) in(a[i]);
    bs_ans();
    return 0;
}