题目链接:
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
", 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;
}