bzoj 2298 [HAOI2011]problem a dp+树状数组

题面

题目传送门

解法

考虑补集转化，我们只要求正确的最大个数即可

显然，一些明显就是错误的东西可以直接排除

对于\((x,y)\)相同的位置一定相等

那么我们就可以把\((x,y)\)相等的并在一类

然后考虑一下\((x,y)\)怎么转化，显然就是那一个数在整个数列中排名对应的区间，为\([x+1,n-y]\)

把区间相等的合并一下，剩下的都是不相等的

然后只要选出尽量多的不相交的区间即可

将左端点排序，然后用树状数组优化一下dp即可

时间复杂度：\(O(n\ log\ n)\)

代码

#include <bits/stdc++.h>
#define N 100010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
	int l, r, v;
	bool operator < (const Node &a) const {
		if (l != a.l) return l < a.l;
		return r < a.r;
	}
} a[N], b[N];
int n, tot, len, f[N], dp[N];
int lowbit(int x) {return x & -x;}
bool cmp(Node a, Node b) {return a.l != b.l || a.r != b.r;}
int query(int x) {
	int ret = 0;
	for (int i = x; i; i -= lowbit(i))
		chkmax(ret, f[i]);
	return ret;
}
void modify(int x, int v) {
	for (int i = x; i <= n; i += lowbit(i))
		chkmax(f[i], v);
}
int main() {
	read(n);
	for (int i = 1; i <= n; i++) {
		int x, y; read(x), read(y);
		if (x + y < n) a[++tot] = (Node) {x + 1, n - y, 0};
	}
	sort(a + 1, a + tot + 1);
	Node las = a[1]; int sum = 1;
	for (int i = 2; i <= tot; i++)
		if (cmp(a[i], las)) {
			b[++len] = las, b[len].v = sum;
			las = a[i], sum = 1;
		} else sum++;
	b[++len] = las; b[len].v = sum; int ans = 0;
	for (int i = 1; i <= len; i++) chkmin(b[i].v, b[i].r - b[i].l + 1);
	sort(b + 1, b + len + 1);
	for (int i = 1; i <= len; i++) {
		dp[i] = query(b[i].l - 1) + b[i].v;
		modify(b[i].r, dp[i]); chkmax(ans, dp[i]);
	}
	cout << n - ans << "\n";
	return 0;
}