2019年牛客多校第一场 I题Points Division 线段树+DP
题目链接
传送门
题意
给你 n n n个点，每个点的坐标为 ( x i , y i ) (x_i,y_i) (xi,yi)，有两个权值 a i , b i a_i,b_i ai,bi。
现在要你将它分成 A , B \mathbb{A},\mathbb{B} A,B两部分，使得在满足“ A \mathbb{A} A的点不能落在在 B \mathbb{B} B的点的右下方”的条件下 ∑ i ∈ A a i + ∑ j ∈ B b j \sum\limits_{i\in\mathbb{A}}a_i+\sum\limits_{j\in\mathbb{B}}b_j i∈A∑ai+j∈B∑bj最大。
思路
这篇博客讲得很详细，大家可以看这位大佬的昂~
代码实现如下
#include
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
int n;
vector<int> vec;
struct Point {
int x, y, a, b;
bool operator < (const Point& pp) const {
return x == pp.x ? y > pp.y : x < pp.x;
}
}point[maxn];
struct node {
int l, r;
LL mx, lazy;
}segtree[maxn<<2];
void push_up(int rt) {
segtree[rt].mx = max(segtree[lson].mx, segtree[rson].mx);
}
void push_down(int rt) {
LL x = segtree[rt].lazy;
segtree[rt].lazy = 0;
segtree[lson].lazy += x;
segtree[rson].lazy += x;
segtree[lson].mx += x;
segtree[rson].mx += x;
}
void build(int rt, int l, int r) {
segtree[rt].l = l, segtree[rt].r = r;
segtree[rt].mx = segtree[rt].lazy = 0;
if(l == r) return;
int mid = (segtree[rt].l + segtree[rt].r) >> 1;
build(lson, l, mid);
build(rson, mid + 1, r);
}
void update1(int rt, int pos, LL val) {
if(segtree[rt].l == segtree[rt].r) {
segtree[rt].mx = val;
return;
}
push_down(rt);
int mid = (segtree[rt].l + segtree[rt].r) >> 1;
if(pos <= mid) update1(lson, pos, val);
else update1(rson, pos, val);
push_up(rt);
}
void update2(int rt, int l, int r, LL val) {
if(segtree[rt].l == l && segtree[rt].r == r) {
segtree[rt].mx += val;
segtree[rt].lazy += val;
return;
}
push_down(rt);
int mid = (segtree[rt].l + segtree[rt].r) >> 1;
if(r <= mid) update2(lson, l, r, val);
else if(l > mid) update2(rson, l, r, val);
else {
update2(lson, l, mid, val);
update2(rson, mid + 1, r, val);
}
push_up(rt);
}
LL query(int rt, int l, int r) {
if(segtree[rt].l == l && segtree[rt].r == r) {
return segtree[rt].mx;
}
push_down(rt);
int mid = (segtree[rt].l + segtree[rt].r) >> 1;
if(r <= mid) return query(lson, l, r);
else if(l > mid) return query(rson, l, r);
else return max(query(lson, l, mid), query(rson, mid + 1, r));
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
while(~scanf("%d", &n)) {
vec.clear();
for(int i = 1; i <= n; ++i) {
scanf("%d%d%d%d", &point[i].x, &point[i].y, &point[i].a, &point[i].b);
vec.push_back(point[i].y);
}
sort(vec.begin(), vec.end());
vec.erase(unique(vec.begin(), vec.end()), vec.end());
sort(point + 1, point + n + 1);
for(int i = 1; i <= n; ++i) {
point[i].y = lower_bound(vec.begin(), vec.end(), point[i].y) - vec.begin() + 1;
}
int sz = vec.size();
build(1, 0, sz + 1);
for(int i = 1; i <= n; ++i) {
LL num = query(1, 0, point[i].y);
update1(1, point[i].y, num + point[i].b);
update2(1, 0, point[i].y - 1, point[i].a);
update2(1, point[i].y + 1, sz + 1, point[i].b);
}
printf("%lld\n", segtree[1].mx);
}
return 0;
}