19暑假dp优化A

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 ti at section ai. If you are at section x (1 ≤ x ≤ n) at the time of i-th launching, you’ll gain happiness value bi - |ai - 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 ai, bi, ti (1 ≤ ai ≤ n; 1 ≤ bi ≤ 109; 1 ≤ ti ≤ 109). The i-th line contains description of the i-th launching.

It is guaranteed that the condition ti ≤ ti + 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
Input
50 3 1
49 1 1
26 1 4
6 1 10
Output
-31
Input
10 2 1
1 1000 4
9 1000 4
Output
1992
dp单调队列优化板题。。。今天光写板题了。。。

#include<bits/stdc++.h> 
#define ll long long
const int maxn=1e7+10;
using namespace std;
ll n,m,d;
struct node
{
    ll a,b,t;
}p[301];
ll dp[2][150001];
ll q[150001];
ll cmp(node n1,node n2)//根据烟花时间排序，小的在前
{
    return n1.t<n2.t;
}
int main()
{
    ll time;//上一个的时间
    scanf("%I64d%I64d%I64d",&n,&m,&d);
    for(ll i=1;i<=m;i++)
    scanf("%I64d%I64d%I64d",&p[i].a,&p[i].b,&p[i].t);
    sort(p+1,p+1+m,cmp);
    memset(dp,0,sizeof(dp));
    time=p[1].t;
    ll mm=0;//滚动数组，1-mm为上一次的数组 
    for(ll i=1;i<=m;i++)
    {
        ll l=0,r=0,k=1;
        if(time==p[i].t)//如果时间相同则直接更新 
        {
            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].t-time;
            time=p[i].t;
            for(ll j=1;j<=n;j++)
            {
                while(k<=n&&k<=j+d*t)           //从大到小遍历一次得到一个从大到小的队列 
                {
                    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])  //在范围中选出第一个即最大的 
                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("%I64d\n",ans);
}