POJ 2482 Stars in Your Window（线段树）

POJ 2482 Stars in Your Window

题目链接

题意：给定一些星星，每个星星都有一个亮度，现在要用w * h的矩形去框星星，问最大能框的亮度是多少

思路：转化为扫描线的问题，每个星星转化为一个矩形，那么等于求矩形相交区域值最大的区域，利用线段树去维护即可

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 10005;

int n, w, h;

struct Line {
ll l, r, y, c;
Line() {}
Line(ll l, ll r, ll y, ll c) {
this->l = l; this->r = r;
this->y = y; this->c = c;
}
} line[N * 2];

bool cmp(Line a, Line b) {
if (a.y == b.y) return a.c < b.c;
return a.y < b.y;
}

ll hash[N * 2];
int hn;

int find(ll x) {
return lower_bound(hash, hash + hn, x) - hash;
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)

struct Node {
int l, r;
ll add, sum;
} node[N * 8];

void build(int l, int r, int x = 0) {
node[x].l = l; node[x].r = r;
node[x].add = node[x].sum = 0;
if (l == r) return;
int mid = (l + r) / 2;
build(l, mid, lson(x));
build(mid + 1, r, rson(x));
}

void pushup(int x) {
if (node[x].l == node[x].r) node[x].sum = node[x].add;
else node[x].sum = max(node[lson(x)].sum, node[rson(x)].sum) + node[x].add;
}

void add(int l, int r, ll c, int x = 0) {
if (node[x].l >= l && node[x].r <= r) {
node[x].add += c;
pushup(x);
return;
}
int mid = (node[x].l + node[x].r) / 2;
if (l <= mid) add(l, r, c, lson(x));
if (r > mid) add(l, r, c, rson(x));
pushup(x);
}

int main() {
while (~scanf("%d%d%d", &n, &w, &h)) {
ll x1, y1, x2, y2, c;
for (int i = 0; i < n; i++) {
scanf("%lld%lld%lld", &x1, &y1, &c);
x2 = x1 + w; y2 = y1 + h;
hash[i * 2] = x1; hash[i * 2 + 1] = x2;
line[i * 2] = Line(x1, x2, y1, c);
line[i * 2 + 1] = Line(x1, x2, y2, -c);
}
n *= 2;
hn = 1;
sort(hash, hash + n);
for (int i = 1; i < n; i++)
if (hash[i] != hash[i - 1])
hash[hn++] = hash[i];
build(0, hn - 2);
sort(line, line + n, cmp);
ll ans = 0;
for (int i = 0; i < n; i++) {
add(find(line[i].l), find(line[i].r) - 1, line[i].c);
ans = max(ans, node[0].sum);
}
printf("%lld\n", ans);
}
return 0;
}