POJ 2482 Stars in Your Window

题目地址
题意：这是我看过最浪漫的题目，其实题意特别简单，就是有一扇窗户长w宽h，然后天上有n个星星，告诉你星星的位置和星星的亮度，让你求出你能透过窗户看到的星星的总亮度最大是多少
思路：我看到这道题的时候想到了cf前段时间一次比赛的题目，感觉蛮像的，当时我还是想用线段树写但是那天TLE，我看到这题一看到数据我就发现肯定那种办法还是TLE的，参考别人的博客才发现这道题可以转化为扫描线去写，每个矩形的价值都是星星亮度，这样就可以找到最大的星星亮度之和。（怎么处理边界呢？我的写法是把长度-1，保证区间端点重复问题不会出现，宽度那里就是排序的时候先让出边先被查找到）

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack> #include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 50010#define LL __int64#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1e9 + 7;const double eps = 1e-8;const double pi = acos(-1.0);struct nope {    LL x1, x2, y, k;}num[N];bool cmp(nope a, nope b) {    if (a.y == b.y) {        return a.k < b.k;    }    return a.y < b.y;}LL sum[N << 2];LL lazy[N << 2];vector<LL> v;struct Segment__Tree {    int x, y;    void pushUp(int ans) {        sum[ans] = max(sum[movel], sum[mover]) + lazy[ans];//这个是我疏忽的点，忘了一个标记，因为是最大值没法把值传递到下面去所以，可能会被覆盖    }    void build(int l, int r, int ans) {        memset(sum, 0, sizeof(sum));    }    void updata(int l, int r, int ans, LL nums) {        if (l >= x&&r <= y) {            sum[ans] += nums;            lazy[ans] += nums;            return;        }        int mid = getMid;        if (mid<x) {            updata(rson, nums);        }        else if (mid >= y) {            updata(lson, nums);        }        else {            updata(lson, nums);            updata(rson, nums);        }        pushUp(ans);    }};int main() {    cin.sync_with_stdio(false);    Segment__Tree tree;    LL n, w, h;    LL a, b, c;    while (cin >> n >> w >> h) {        v.clear();        w--;        for (int i = 0; i < n; i++) {            cin >> a >> b >> c;            num[2 * i].k = c;            num[2 * i].x1 = a;            num[2 * i].x2 = a + w;            num[2 * i].y = b;            num[2 * i + 1].k = -c;            num[2 * i + 1].x1 = a;              num[2 * i + 1].x2 = a + w;            num[2 * i + 1].y = b + h;            v.push_back(a);            v.push_back(a + w);        }        if (w == -1 || h == -1) {            cout << 0 << endl;            continue;        }        sort(v.begin(), v.end());        int len = unique(v.begin(), v.end()) - v.begin();        tree.build(1, len, 1);        LL mmax = -1;        n *= 2;        sort(num, num + n, cmp);        for (int i = 0; i < n; i++) {            tree.x = lower_bound(v.begin(), v.begin() + len, num[i].x1) - v.begin() + 1;            tree.y = lower_bound(v.begin(), v.begin() + len, num[i].x2) - v.begin() + 1;            tree.updata(1, len, 1, num[i].k);            if (num[i].k > 0) {                mmax = max(mmax, sum[1]);            }        }        cout << mmax << endl;    }    return 0;}