CF377D Developing Game

题目链接：

luogu

题目分析：

把每个人当成一个三元组\([l_i, r_i, v_i]\)
考虑每个人对哪个能力区间\([L, R]\)有贡献
应该是左端点在\([l_i, v_i]\)，右端点在\([v_i, r_i]\)的区间
拍到一个二维平面上，求最多有多少个矩形交一起
线段树维护扫描线即可

代码：

#include<bits/stdc++.h>
#define N (600000 + 10)
using namespace std;
inline int read() {
    int cnt = 0, f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
    while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();}
    return cnt * f;
}

int n, Y[N], cnt, ans = 0, ansx, ansy, tot;
int l[N], r[N], v[N];
struct node2 {
    int x, Y1, Y2, tag;
}seg[N << 1];

struct node {
    int l, r;
    int tag, gmax, pos;
    #define l(p) tree[p].l
    #define r(p) tree[p].r
    #define tag(p) tree[p].tag
    #define gmax(p) tree[p].gmax
    #define pos(p) tree[p].pos
} tree[N << 2];

void pushdown(int p) {
    tag(p << 1) += tag(p);
    tag(p << 1 | 1) += tag(p);
    gmax(p << 1) += tag(p);
    gmax(p << 1 | 1) += tag(p);
    tag(p) = 0;
}

void pushup(int p) {
    if (gmax(p << 1) > gmax(p << 1 | 1)) {
        pos(p) = pos(p << 1);
        gmax(p) = gmax(p << 1);
    } else {
        pos(p) = pos(p << 1 | 1);
        gmax(p) = gmax(p << 1 | 1);
    }
}

void build (int p, int l, int r) {
    l(p) = l, r(p) = r;
    if (l == r) {pos(p) = Y[l]; return;}
    int mid = (l + r) >> 1;
    build (p << 1, l, mid);
    build (p << 1 | 1, mid + 1, r);
    pos(p) = pos(p << 1);
}

void modify(int p, int l, int r, int k) {
    pushdown(p);
    if (l <= Y[l(p)] && r >= Y[r(p)]) {tag(p) += k, gmax(p) += k; return;}
    int mid = (l(p) + r(p)) >> 1;
    if (l <= Y[mid]) modify(p << 1, l, r, k);
    if (r > Y[mid]) modify(p << 1 | 1, l, r, k);
    pushup(p);
}
bool cmp(node2 a, node2 b) {
    return a.x == b.x ? a.tag > b.tag : a.x < b.x;
}

int main() {
//  freopen("data.in", "r", stdin);
//  freopen("myself.out", "w", stdout);
    n = read();
    for (register int i = 1; i <= n; ++i) {
        l[i] = read(), v[i] = read(), r[i] = read();
        seg[++tot].x = l[i], seg[tot].Y1 = v[i], seg[tot].Y2 = r[i], seg[tot].tag = 1, Y[tot] = v[i];
        seg[++tot].x = v[i], seg[tot].Y1 = v[i], seg[tot].Y2 = r[i], seg[tot].tag = -1, Y[tot] = r[i];
    }
    sort (Y + 1, Y + tot + 1);
    for (register int i = 1; i <= tot; ++i) if (Y[i] != Y[i + 1]) Y[++cnt] = Y[i];
    build (1, 1, cnt);
    sort (seg + 1, seg + tot + 1, cmp);
//  for (register int i = 1; i <= cnt; ++i) cout<<Y[i]<<" "; return 0;
    for (register int i = 1; i <= tot; ++i) {
        modify(1, seg[i].Y1, seg[i].Y2, seg[i].tag);
//      cout<<gmax(1)<<" ";
        if (gmax(1) > ans) {
            ans = gmax(1);
            ansx = seg[i].x;
            ansy = pos(1);
        }
    }
//  return 0;
    printf("%d\n", ans);
//  cout<<ansx<<" "<<ansy<<endl; return 0;
    for (register int i = 1; i <= n; ++i) 
        if (v[i] <= ansy && r[i] >= ansy && l[i] <= ansx && v[i] >= ansx) printf("%d ", i);
    return 0;
}

来源：https://www.cnblogs.com/kma093/p/11791979.html