BZOJ 2131 免费的馅饼

${color{cyan}{>>Qusetion}}$

首先看到每个物品(馅饼)有三个属性(时间$t$,位置$p$,价值$v$),容易想到以物品为阶段来$dp$

令$f[i]$表示到第$i$个物品的最大收益

有
$$f[i] = maxleft { f[j] ight }+v[i],2*(t[i] - t[j])geq left | p[i] - p[j] ight |$$

将$t[i]*2$,然后将限制条件化简可得

$$left{egin{matrix}
t[j]-p[j]leq t[i]-p[i]\
t[j]+p[j]leq t[i]+p[i]
end{matrix} ight.$$

令$t[i]-p[i] = x,t[i]+p[i] = y$

就成了模板的二维偏序,第一维排序,第二维树状数组维护即可(与P3431 [POI2005]AUT-The Bus一模一样)

代码如下

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#define ll long long
using namespace std; 

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//-------------------------------------------------------

const int N = 1e5+7;

int n,m;
ll b[N],ans;

struct obj {
    ll x,y;int v;
    bool operator < (const obj &sed) const {
        return x == sed.x ? y < sed.y : x < sed.x;
    } 
}p[N];

struct Map {
    ll pos,v;
    bool operator < (const Map &sed) const {
        return v == sed.v ? pos < sed.pos : v < sed.v;
    }
}a[N];

void ls_y() {
    sort(a+1,a+n+1);
    int _id = 0,i;
    for(i = 1;i <= n; ++i) {
        if(a[i].v != a[i-1].v) ++_id;
        p[a[i].pos].y = _id;
    }
}

ll Q(int pos) {
    ll res = 0;
    for(int i = pos;i;i -= i&-i) res = max(res,b[i]);
    return res;
}

void A(int pos,ll k) {
    for(int i = pos;i <= n;i += i&-i) b[i] = max(b[i],k);
}

int main() {
    //freopen("0.in","r",stdin);
    int i; ll t,pos; in(m); in(n);
    for(i = 1;i <= n; ++i) {
        in(t); in(pos); in(p[i].v);
        p[i].x = (t<<1) - pos; p[i].y = (t<<1) + pos;
        a[i].pos = i,a[i].v = p[i].y;
    }
    ls_y();
    sort(p+1,p+n+1);
    for(i = 1;i <= n; ++i) {
        ll _f = Q(p[i].y) + p[i].v;
        ans = max(ans,_f);
        A(p[i].y,_f);
    }
    out(ans);
    return 0;
}