loj 10181 绿色通道

分析

最大值最小，考虑二分

当状态太多，就可以考虑二分分掉一个状态；

剩下的就好办了，不过注意边界&初值

代码

/***********************
User:Mandy.H.Y
Language:c++
Problem:loj 10181
Algorithm: 
***********************/

#include<bits/stdc++.h>

using namespace std;

const int maxn = 5e4 + 5;

int n,t;
int a[maxn],q[maxn];
int dp[maxn][3];

template<class T>inline void read(T &x){
    x = 0;bool flag = 0;char ch = getchar();
    while(!isdigit(ch)) flag |= ch == '-',ch = getchar();
    while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48),ch = getchar();
    if(flag) x = -x;
}

template<class T>void putch(const T x){
    if(x > 9) putch(x / 10);
    putchar(x % 10 | 48);
}

template<class T>void put(const T x){
    if(x < 0) putchar('-'),putch(-x);
    else putch(x);
}

void file(){
    freopen("10181.in","r",stdin);
    freopen("10181.out","w",stdout);
}

void readdata(){
    read(n);read(t);
    for(int i = 1;i <= n; ++ i) read(a[i]);
}

bool check(int x){
    int l,r;
    l = r = 0;
    dp[0][0] = 0;
    dp[0][1] = 0;//初始化 
    for(int i = 1;i <= n; ++ i){
        while(l < r && q[l] < i - x) l++;
        dp[i][1] = min(dp[i-1][0],dp[i-1][1]) + a[i];
        if(l < r) dp[i][0] = dp[q[l]][1];
        if(i <= x) dp[i][0] = 0;//注意初值 
        while(l < r && dp[q[r-1]][1] >= dp[i][1]) r--;
        q[r++]=i;
    }
    int ans = min(dp[n][0],dp[n][1]);
    if(ans <= t) return 1;
    else return 0;
}

void work(){
    int l = 0,r = n,ans = 0;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(check(mid)) ans = mid,r = mid - 1;
        else l = mid + 1;
    } 
    put(ans);
}

int main(){
//    file();
    readdata();
    work();
    return 0;
}