Title Description

The streets of Byte City form a regular, chessboardlike network - they are either north-south or west-east directed. We shall call them NS- and WE-streets. Furthermore, each street crosses the whole city. Every NS-street intersects every WE- one and vice versa. The NS-streets are numbered from \(1\) to \(n\), starting from the westernmost. The WE-streets are numbered from \(1\) to \(m\), beginning with the southernmost. Each intersection of the \(i\)'th NS-street with the \(j\)'th WE-street is denoted by a pair of numbers \((i,j)\) (for \(1\le i\le n\), \(1\le j\le m\)).

There is a bus line in Byte City, with intersections serving as bus stops. The bus begins its itinerary by the \((1,1)\) intersection, and finishes by the \((n,m)\) intersection. Moreover, the bus may only travel in the eastern and/or northern direction.

There are passengers awaiting the bus by some of the intersections. The bus driver wants to choose his route in a way that allows him to take as many of them as possible. (We shall make an assumption that the interior of the bus is spacious enough to take all of the awaiting passengers, regardless of the route chosen.)TaskWrite a programme which:

reads from the standard input a description of the road network and the number of passengers waiting at each intersection,finds, how many passengers the bus can take at the most,writes the outcome to the standard output.

Byte City streets form a checkerboard network standard - they are either north to south or east to west is the North-South direction of the intersection from 1 to n number of road west trending number from 1 to m each intersection with. two numbers (i, j) represents (1 <= i <= n, 1 <= j <= m). Byte City there is a bus line, provided in some of the junction bus stops. from the bus (1, 1) start, the (n, m) end. buses can only go east or north. there are some sites, some passengers waiting for the bus. bus drivers want route can receive as many passengers. help think about how he can receive the most passengers.

Input Format

The first line of the standard input contains three positive integers \(n\), \(m\) and \(k\) - denoting the number of NS-streets, the number of WE-streets and the number of intersections by which the passengers await the bus, respectively \((1\le n\le 10^9, 1\le m\le 10^9, 1\le k\le 10^5)\).

The following \(k\) lines describe the deployment of passengers awaiting the bus, a single line per intersection. In the \((i+1)\)'st line there are three positive integers \(x_i, y_i\) and \(p_i\), separated by single spaces, \(1\le x_i\le n,1\le y_i\le m,1\le p_i\le 10^6\) . A triplet of this form signifies that by the intersection\((x_i,y_i)p_i\) passengers await the bus. Each intersection is described in the input data once at the most. The total number of passengers waiting for the bus does not exceed \(1\ 000\ 000\ 000\).

Output Format

Your programme should write to the standard output one line containing a single integer - the greatest number of passengers the bus can take.

Sample input and output

Entry

Export

Thinking

A first thought \ (n \ times m \) of the DP, but because \ (n, m \) are \ (10 ^ 9 \) , it is certainly not

It may be noted, though \ (n, m \) is large, but the number of points but rarely, and only \ (10 ^ 5 \) months, so you can consider the first point discrete, so time is from \ (O (n \ times m) down to O (K ^ 2) \) , but will still timeout

In this case, we can treat each tap abscissa ascending order, if the same abscissa, ordinate sorted in ascending order, then DP

State transition equation: \ (DP [I] = max (DP [. 1], DP [2] DP ... [I]) + P [I] \) for the \ (max (DP [I]) \) , Fenwick tree can be used to seek

Code

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=2e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
struct wzy
{
    int x,y,s;
}p[maxn];
int c[maxn];
int mapx[maxn],mapy[maxn];
bool cmp(wzy u,wzy v)
{
    if(u.x==v.x)
        return u.y<v.y;
    return u.x<v.x;
}
int lowbit(int x)
{
    return x&(-x);
}
void update(int place,int num,int n)
{
    while(place<=n)
    {
        c[place]=max(c[place],num);
        place+=lowbit(place);
    }
}
int query(int place)
{
    int ans=0;
    while(place>0)
    {
        ans=max(ans,c[place]);
        place-=lowbit(place);
    }
    return ans;
}
int dp[maxn];
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in.txt", "r", stdin);
        freopen("/home/wzy/out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m,k;
    cin>>n>>m>>k;
    for(int i=1;i<=k;i++)
    {
        cin>>p[i].x>>p[i].y>>p[i].s;
        mapx[i]=p[i].x;
        mapy[i]=p[i].y;
    }
    // 离散化
    sort(mapx+1,mapx+1+k);
    sort(mapy+1,mapy+1+k);
    int numx,numy;
    numx=numy=k;
    numx=unique(mapx+1,mapx+1+numx)-(mapx+1);
    numy=unique(mapy+1,mapy+1+numy)-(mapy+1);
    for(int i=1;i<=k;i++)
    {
        p[i].x=lower_bound(mapx+1,mapx+numx+1,p[i].x)-mapx;
        p[i].y=lower_bound(mapy+1,mapy+numy+1,p[i].y)-mapy;
    }
    sort(p+1,p+1+k,cmp);
    for(int i=1;i<=k;i++)
    {
        dp[i]=query(p[i].y)+p[i].s;
        update(p[i].y,dp[i],k);
    }
    int ans=0;
    for(int i=1;i<=k;i++)
        ans=max(ans,dp[i]);
    cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}

Luo Valley P3431: [POI2005] AUT-The Bus (+ DP + tree discrete array)