2021--ICPC网络预选第一场--A Busiest Computing Nodes

比赛时思路不正确，赛后看大佬的代码才知道是线段树+二分。

这题和D题一样，维护的都是区间最小值，当时肝出了D，要是用D的思路来想想这题，说不定也能做出来了。

线段树维护的是解决完问题的时间的区间最小值，所以区间的值是单调递减的，我们想在[i%k,k-1]中找一个<=start的最靠左的值，可以用二分来优化遍历数组。如果在[i%k,k-1]中找不着，那就只好去[0,i%k-1]里面找了，再找不着，只好放弃了。

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5 + 5, INF = 1e18;
int tree[N * 4];
int book[N];
int x, y, val;
void up(int now, int l, int r)
{if (l == r) {tree[now] = val;return;}int chl = 2 * now, chr = 2 * now + 1;int mid = (l + r) >> 1;if (x <= mid) up(chl, l, mid);else up(chr, mid + 1, r);tree[now] = min(tree[chl], tree[chr]);
}
int que(int now, int l, int r)
{if (x <= l && y >= r) {return tree[now];}int chl = 2 * now, chr = 2 * now + 1;int mid = (l + r) >> 1;int ret = INF;if (x <= mid) ret = min(ret, que(chl, l, mid));if (y > mid) ret = min(ret, que(chr, mid + 1, r));return ret;
}
signed main()
{int k, n;scanf("%lld%lld", &k, &n);for (int i = 0; i < n; i++) {int s, e;scanf("%lld%lld", &s, &e);int l = i % k, r = k - 1;while(l < r) {int mid = (l + r) >> 1;x = l, y = mid;if (que(1, 0, k - 1) <= s) r = mid;else l = mid + 1;}x = l, y = l;if (que(1, 0, k - 1) <= s) {x = l, val = s + e;book[l]++;up(1, 0, k - 1);}else {l = 0, r = i % k - 1;while(l < r) {int mid = (l + r) >> 1;x = l, y = mid;if (que(1, 0, k - 1) <= s) r = mid;else l = mid + 1;}x = l, y = l;if (que(1, 0, k - 1) <= s) {x = l, val = s + e;book[l]++;up(1, 0, k - 1);}}}int ma = 0;for (int i = 0; i < k; i++) {ma = max(ma, book[i]);}vector<int> ans;for (int i = 0; i < k; i++) {if (ma == book[i]) ans.push_back(i);}for (int i = 0; i < ans.size(); i++) {printf("%lld", ans[i]);if (i != ans.size() - 1) printf(" ");}return 0;
}

顺便附上D题的线段树，这题比起A就是多了个懒标记，下传即可。。
D Edge of Taixuan

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5 + 5, INF = 0x3f3f3f3f;
int tree[N * 4], add[N * 4];
int x, y, val;
void down(int now)
{if (add[now] == INF) return;int chl = 2 * now, chr = 2 * now + 1;tree[chl] = min(tree[chl], add[now]);tree[chr] = min(tree[chr], add[now]);add[chl] = min(add[chl], add[now]);add[chr] = min(add[chr], add[now]);add[now] = INF;
}
void up(int now, int l, int r)
{if (x <= l && y >= r) {tree[now] = min(tree[now], val);add[now] = min(add[now], val);return;}down(now);int chl = 2 * now, chr = 2 * now + 1;int mid = (l + r) >> 1;if (x <= mid) up(chl, l, mid);if (y > mid) up(chr, mid + 1, r);tree[now] = min(tree[chl], tree[chr]);
}
int que(int now, int l, int r)
{if (x <= l && y >= r) {return tree[now];}down(now);int chl = 2 * now, chr = 2 * now + 1;int mid = (l + r) >> 1;int ret = INF;if (x <= mid) ret = min(ret, que(chl, l, mid));if (y > mid) ret = min(ret, que(chr, mid + 1, r));return ret;
}
signed main()
{int t;scanf("%lld", &t);int cnt = 0;while(t--) {memset(tree, INF, sizeof(tree));memset(add, INF, sizeof(add));int n, m;scanf("%lld%lld", &n, &m);int sum = 0;for (int i = 1; i <= m; i++) {scanf("%lld%lld%lld", &x, &y, &val);sum += (y - x) * (y - x + 1) / 2 * val;y--;up(1, 1, n - 1);}// cout << "sum = " << sum << endl;int ans = 0;bool ck = 1;for (int i = 1; i <= n - 1; i++) {x = y = i;int ret = que(1, 1, n - 1);if (ret == INF) {ck = 0;break;}ans += ret;}if (ck) printf("Case #%lld: %lld", ++cnt, sum - ans);else printf("Case #%lld: Gotta prepare a lesson", ++cnt);if (t) printf("\n");}return 0;
}

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