codeforces 573B B. Bear and Blocks(线段树+dp)

题目链接：

codeforces 573B

题目大意：

给出n个连续塔，每个塔有高度$h_i$,每次取走最外层的块，问需要多少次操作能够拿光所有的块。

题目分析：

• 首先我们可以知道第一次操作时，对于每个塔的变化满足如下的公式： $$h_i =min ( h_{i-1} , h_i -1 , h_{i+1})$$
• 每次操作都满足如下的递推式，我们递推一下得到第k次操作第i的塔的高度： $$h_i = max ( 0 , \min_{j=0}^i \{ min( h_{i-j}-(k-j) , h_{i+j}-(k-j) \} \ \Rightarrow h_i = max ( 0 , \min_{j=0}^i \{ min( h_{i-j}+j-k , h_{i+j}+j-k \}$$
  因为k是常数，所以对于，每一项我们找到的就是最小的$h_{i-j}+j和h_{i+j}+j$，这可用线段树进行维护。
• 而且要设定两个虚拟的高度为0的塔，分别放在这一堆塔的两头，这样能够维护两侧的在每次操作拿光的情况。
• 至于线段树维护的部分，就是遍历i两次，分别从小到大和从大到小，每次修改之前遍历过的位置为所要的状态，也就是之前遍历过的位置区间内全部加1，然后统计[0,i]的最小值，右侧的同理。
  

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AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <set>
#define MAX 100007

using namespace std;

int n;
int l[MAX],r[MAX],h[MAX];

struct Tree
{
    int l,r,minn,lazy;
}tree[MAX<<2];

void push_up ( int u )
{
    tree[u].minn = min ( tree[u<<1].minn , tree[u<<1|1].minn );
}

void build ( int u , int l , int r )
{
    tree[u].l = l;
    tree[u].r = r;
    tree[u].lazy = 0;
    int mid = l+r>>1;
    if ( l == r ) 
    {
        tree[u].minn = h[l];
        return;
    }
    build ( u<<1 , l , mid );
    build ( u<<1|1 , mid+1 , r );
    push_up ( u );
}

void push_down ( int u )
{
    int ll = tree[u].lazy;
    tree[u].lazy = 0;
    if ( ll )
    {
        tree[u<<1].lazy += ll;
        tree[u<<1|1].lazy += ll;
        tree[u<<1].minn += ll;
        tree[u<<1|1].minn += ll;
    }
}

void update ( int u , int left , int right )
{
    int l = tree[u].l;
    int r = tree[u].r;
    int mid = l+r>>1;
    if ( left <= l && r <= right )
    {
        tree[u].minn += 1;
        tree[u].lazy += 1;
        return;
    }
    push_down ( u );
    if ( left <= mid && right >= l ) 
        update ( u<<1 , left , right );
    if ( left <= r && right > mid )
        update ( u<<1|1 , left , right );
    push_up ( u );
}

int query ( int u , int left , int right )
{
    int l = tree[u].l;
    int r = tree[u].r;
    int mid = l+r>>1;
    if ( left <= l && r <= right )
        return tree[u].minn;
    push_down ( u );
    int ret = 1<<29;
    if ( left <= mid && right >= l ) 
        ret = min ( ret , query ( u<<1 , left , right ) );
    if ( left <= r && right > mid )
        ret = min ( ret , query ( u<<1|1 , left , right ) );
    return ret;
}

int main ( )
{
    while ( ~scanf ( "%d" , &n ) )
    {
        h[0] = h[n+1] = 0;
        for ( int i = 1; i <= n ; i++ )
            scanf ( "%d" , &h[i] );
        build ( 1 , 0 , n+1 );
        for ( int i = 1 ; i <= n ; i++ )
        {
            update ( 1 , 0 , i-1 );
            l[i] = query ( 1 , 0 , i );
        }
        build ( 1 , 0 , n+1 );
        for ( int i = n ; i > 0 ; i-- )
        {
            update ( 1 , i+1 , n+1 );
            r[i] = query ( 1 , i , n+1 );
        }
        int ans = 0;
        for ( int i = 1 ; i <= n ; i++ )
        {
            //cout << "YES : " << l[i] << " " << r[i] << endl;
            ans = max ( ans , min ( l[i] , r[i] ));
        }
        printf ( "%d\n" , ans );
    }
}