hihoCoder 1079 离散化（线段树离散化）

来源：互联网发布：华为网络认证多少钱编辑：程序博客网时间：2024/05/17 07:47

根据提示我们要给长度为L的区间建立线段树，但是L很大，所以不能直接对L建树。

转化一下就是一共n个区间，最多2*n个点，所以我们可以使用较小的2*n个数来代替这些比较大的数而保证他们的相对大小不发生变化。然后我们就可以对长度为2*n的区间建立线段树了。

然后需要注意的一点是离散的线段树结点为[i,i]，左儿子区间为[l,mid]，右儿子区间为[mid+1,r]。

但是对于连续的区间线段树结点为[i,i+1]，左儿子区间为[l,mid]，右儿子区间为[mid,r]

然后按照线段树的方法做就Ok了。

#pragma warning(disable:4996)#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 100005;const int M = (N << 2) * 2;int l[M], r[M], lazy[M], num[M];//num为最大值int a[N], b[N], c[2 * N], id[2 * N];void pushUp(int i){num[i] = max(num[i * 2], num[i * 2 + 1]);}void pushDowm(int i){if (lazy[i] == 0)return;lazy[i * 2] = lazy[i];lazy[i * 2 + 1] = lazy[i];num[i * 2] = num[i * 2 + 1] = lazy[i];lazy[i] = 0;}void build(int ll, int rr, int i){l[i] = ll, r[i] = rr;if (ll + 1 == rr){num[i] = 0;return;}int mid = (ll + rr) >> 1;build(ll, mid, 2 * i);build(mid, rr, 2 * i + 1);pushUp(i);}void update(int ll, int rr, int id, int i){if (l[i] >= ll&&r[i] <= rr){num[i] = id;lazy[i] = id;return;}if (lazy[i])pushDowm(i);int mid = (l[i] + r[i]) >> 1;if (rr <= mid)update(ll, rr, id, i * 2);else if (ll >= mid)update(ll, rr, id, i * 2 + 1);else{update(ll, mid, id, i * 2);update(mid, rr, id, i * 2 + 1);}pushUp(i);}int query(int ll, int rr, int i){if (ll <= l[i] && rr >= r[i]){return num[i];}if (lazy[i])pushDowm(i);int mid = (l[i] + r[i]) >> 1;if (rr <= mid)return query(ll, rr, i * 2);else if (ll >= mid)return query(ll, rr, i * 2 + 1);else return max(query(ll, mid, i * 2), query(mid, rr, i * 2 + 1));}int main(){int n, L; scanf("%d %d ", &n, &L);for (int i = 1; i <= n; i++){scanf("%d %d", a + i, b + i);c[i] = a[i];c[i + n] = b[i];}sort(c + 1, c + 1 + 2 * n);int cnt = 0;for (int i = 1; i <= 2 * n; i++){if (id[c[i]])continue;id[c[i]] = ++cnt;}build(1, cnt, 1);for (int i = 1; i <= n; i++){int x = id[a[i]], y = id[b[i]];update(x, y, i, 1);}memset(a, 0, sizeof a);for (int i = 1; i < cnt; i++){a[query(i, i + 1, 1)] = 1;}int ans = 0;for (int i = 1; i <= n; i++){ans += a[i];}printf("%d\n", ans);return 0;}

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