RMQ+差分处理（Let Them Slide）Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)

题意：https://codeforc.es/contest/1208/problem/E

现有n行w列的墙,每行有一排连续方块，一排方块可以左右连续滑动，且每个方块都有一个价值，第i 列的价值定义为这列的方块的价值和。求1到w列中每列的最大价值。注：如果一个位置没有方块，那么这个位置的价值为0

思路：

直接RMQ+差分数组即可。

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\草稿.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr
#include <string>
#include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
int abss(int a);
int lowbit(int n);
int Del_bit_1(int n);
int maxx(int a,int b);
int minn(int a,int b);
double fabss(double a);
void swapp(int &a,int &b);
clock_t __STRAT,__END;
double __TOTALTIME;
void _MS(){__STRAT=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef long long ll;
const double E=2.718281828;
const double PI=acos(-1.0);
//const ll INF=(1LL<<60);
const int inf=(1<<30);
const double ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;

int dp[N][35],arr[N];

void rmq_init(int n)
{
    for(int i=1;i<=n;i++)
        dp[i][0]=arr[i];
    for(int j=1;(1<<j)<=n;j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
            dp[i][j]=max(dp[i][j-1],dp[i+(1<<j-1)][j-1]);
}
int rmq(int l,int r)
{
    int k=log2(r-l+1);
    return max(dp[l][k],dp[r-(1<<k)+1][k]);
}
ll ans[N];
ll DP[N];

void solve(int len)
{
    int n;
    sc("%d",&n);
    for(int i=1;i<=n;++i)
        sc("%d",&arr[i]);
    rmq_init(n);
    int L,R;
    int max_=max(0,rmq(1,n));
    if(len>=2*n+1)
        DP[n+1]+=max_,DP[len+1-n]-=max_;
    for(int i=1;i<=n;++i)
    {
        L=max(1,n-(len-i));
        R=min(n,i);
        if(n-(len-i)<1||i>n)
            ans[i]+=max(0,rmq(L,R));
        else
            ans[i]+=rmq(L,R);
    }
    int LL=max(n+1,len-n+1);
    for(int i=LL;i<=len;++i)
    {
        L=max(1,n-(len-i));
        R=min(n,i);
        if(n-(len-i)<1||i>n)
            ans[i]+=max(0,rmq(L,R));
        else
            ans[i]+=rmq(L,R);
    }
}

void PR(ll _[],int n)
{
    for(int i=1;i<=n;++i)
        pr("%lld ",_[i]);
    pr("
");
}
int main()
{
    int n,m;
    sc("%d%d",&n,&m);
    for(int i=1;i<=n;++i)solve(m);//,PR(DP,m);
    for(int i=1;i<=m;++i)
    {
        DP[i]+=DP[i-1];
        pr("%lld ",ans[i]+DP[i]);
    }
//    PR(DP,m);
    return 0;
}

/**************************************************************************************/

int maxx(int a,int b)
{
    return a>b?a:b;
}

void swapp(int &a,int &b)
{
    a^=b^=a^=b;
}

int lowbit(int n)
{
    return n&(-n);
}

int Del_bit_1(int n)
{
    return n&(n-1);
}

int abss(int a)
{
    return a>0?a:-a;
}

double fabss(double a)
{
    return a>0?a:-a;
}

int minn(int a,int b)
{
    return a<b?a:b;
}