题解【LOJ2007】「SCOI2015」国旗计划

题面

考虑如果是一条链怎么做。

其实很简单，按照左端点排序，贪心地能选就选即可。

原题是环，那么断环成链即可。

发现每一次暴力跳会 TLE，于是考虑用倍增优化。

设 (f_{i,j}) 表示从第 (i) 个区间开始选 (2^j) 个区间最远覆盖到哪里。

转移直接倍增即可。

注意算答案时要加上首尾两个区间。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

template <typename T>
inline T gi()
{
	T f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 200003, M = N << 1;

int n, m;
int f[23][M];
int tot, ans[N];
struct Node {int l, r, id;} a[M];

inline bool cmp(Node x, Node y) {return x.l < y.l;}

int main()
{
//	File("");
	n = gi <int> (), m = gi <int> ();
	for (int i = 1; i <= n; i+=1)
	{
		int l = gi <int> (), r = gi <int> ();
		if (l <= r) a[++tot] = {l, r, i}, a[++tot] = {l + m, r + m, i + n};
		else a[++tot] = {l, r + m, i}, a[++tot] = {l + m, m + m, i + n};
	}
	sort(a + 1, a + 1 + tot, cmp);
	int now = 1;
	for (int i = 1; i <= tot; i+=1)
	{
		while (now < tot && a[now + 1].l <= a[i].r) ++now;
		f[0][i] = now;
	}
	for (int j = 1; j <= 21; j+=1)
		for (int i = 1; i <= tot; i+=1)
			f[j][i] = f[j - 1][f[j - 1][i]];
	for (int i = 1; i <= n; i+=1)
	{
		int res = 2, now = i;
		for (int j = 21; j >= 0; j-=1)
			if (f[j][now] && a[f[j][now]].r < a[i].l + m)
				res += (1ll << j), now = f[j][now];
		ans[a[i].id] = res;
	}
	for (int i = 1; i <= n; i+=1) printf("%d ", ans[i]);
	return 0;
}