CF372C Watching Fireworks is Fun(单调队列优化DP)

 

A festival will be held in a town's main street. There are n sections in the main street. The sections are numbered 1 through n from left to right. The distance between each adjacent sections is 1.

In the festival m fireworks will be launched. The i-th (1 ≤ i ≤ m) launching is on time t_i at section a_i. If you are at section x (1 ≤ x ≤ n) at the time of i-th launching, you'll gain happiness value b_i - |a_i - x| (note that the happiness value might be a negative value).

You can move up to d length units in a unit time interval, but it's prohibited to go out of the main street. Also you can be in an arbitrary section at initial time moment (time equals to 1), and want to maximize the sum of happiness that can be gained from watching fireworks. Find the maximum total happiness.

Note that two or more fireworks can be launched at the same time.

Input

The first line contains three integers n, m, d (1 ≤ n ≤ 150000; 1 ≤ m ≤ 300; 1 ≤ d ≤ n).

Each of the next m lines contains integers a_i, b_i, t_i (1 ≤ a_i ≤ n; 1 ≤ b_i ≤ 10⁹; 1 ≤ t_i ≤ 10⁹). The i-th line contains description of the i-th launching.

It is guaranteed that the condition t_i ≤ t_i + 1 (1 ≤ i < m) will be satisfied.

Output

Print a single integer — the maximum sum of happiness that you can gain from watching all the fireworks.

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

 

 

题解：

dp[i][j]表示当前放到了第I支烟花并且放这支烟花的时候他站在j点看。推出状态转移方程：

dp[ i ] [ j ] =max(dp[ i - 1] [ k ]) + b[ i ] - | a[ i ] - j | ,其中  max(1,j-t*d)<=k<=min(n,j+t*d) 

不过每当我选取到 i 个烟花的时候，会先把所有能到他的点都放进单调队列中，那么最优解其实被我们存储了 只要那个最优解和第 i 个烟花的位置满足要求就选他，如果不满足 head++ 直到满足要求

#include<bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
#define pil pair<int,ll>
const ll maxn=1e5+5e4;
using namespace std;
ll n,m,d;
struct node{
    ll a,b,x;
}p[301];
ll dp[2][maxn],q[maxn];
ll cmp(node n1,node n2){return n1.x<n2.x;}
int main()
{
     ll time;
    scanf("%lld%lld%lld",&n,&m,&d);
    for(ll i=1;i<=m;i++)
        scanf("%lld%lld%lld",&p[i].a,&p[i].b,&p[i].x);
    sort(p+1,p+1+m,cmp);
    memset(dp,0,sizeof(dp));
    time=p[1].x;
    ll mm=0;
    for(ll i=1;i<=m;i++)
    {
        ll l=0,r=0,k=1;
        if(time==p[i].x)
        {
            for(ll j=1;j<=n;j++)
                dp[mm][j]=dp[1-mm][j]+p[i].b-abs(p[i].a-j);
        }
        else
        {
            ll t=p[i].x-time;
            time=p[i].x;
            for(ll j=1;j<=n;j++)
            {
                while(k<=n&&k<=j+d*t)
                {
                    while(l<r&&dp[1-mm][k]>=dp[1-mm][q[r-1]]) r--;
                    q[r++]=k++;
                }
                while(l<r&&j-t*d>q[l]) l++;
                ll temp=p[i].b-abs(p[i].a-j);
                dp[mm][j]=dp[1-mm][q[l]]+temp;
            }
        }
        mm=1-mm;
    }
    ll ans=-1e17;
    for(ll i=1;i<=n;i++) ans=max(ans,dp[1-mm][i]);
    printf("%lld\n",ans);
    return 0;
}