POJ 2481 Cows

题目链接：​​传送门​​ 问每条线段被包含了多少次

把线段按左端点排序
左端点相同的按右端点大的在前面
这样就不用考虑左端点的影响了
每次插入一条线段就将1-r加1
询问r-inf的值
因为这时左端点的影响已经忽略

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define
#define

using namespace std;
typedef long long ll;
struct node {
  int l, r, w;
}tree[A];
struct N {
  int l, r, id;
  friend bool operator < (const N a, const N b) {
    return a.l == b.l ? a.r > b.r : a.l < b.l;
  }
}e[A];
void build(int k, int l, int r) {
  tree[k].l = l; tree[k].r = r; tree[k].w = 0;
  if (l == r) return;
  int m = (l + r) >> 1;
  build(k << 1, l, m);
  build(k << 1 | 1, m + 1, r);
}
void change(int k, int pos) {
  if (tree[k].l == tree[k].r) {
    tree[k].w++;
    return;
  }
  int m = (tree[k].l + tree[k].r) >> 1;
  if (pos <= m) change(k << 1, pos);
  else change(k << 1 | 1, pos);
  tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w;
}
int ask(int k, int l, int r) {
  if (tree[k].l >= l and tree[k].r <= r) return tree[k].w;
  int m = (tree[k].l + tree[k].r) >> 1, ans = 0;
  if (l <= m) ans += ask(k << 1, l, r);
  if (r > m) ans += ask(k << 1 | 1, l, r);
  return ans;
}
int n, ans[A];

int main(int argc, char const *argv[]) {
  while (scanf("%d", &n)) {
    if (!n) return 0;
    memset(&tree, 0, sizeof tree);
    for (int i = 1; i <= n; i++) scanf("%d%d", &e[i].l, &e[i].r), e[i].id = i, e[i].l++, e[i].r++;
    build(1, 1, 100000); sort(e + 1, e + n + 1);
    for (int i = 1; i <= n; i++) {
      if (e[i].l == e[i - 1].l and e[i].r == e[i - 1].r)
        ans[e[i].id] = ans[e[i - 1].id];
      else ans[e[i].id] = ask(1, e[i].r, 100000);
      change(1, e[i].r);
    }
    for (int i = 1; i <= n; i++) i == n ? printf("%d\n", ans[i]) : printf("%d ", ans[i]);
  }
  return 0;
}