[POI2014]KAR-Cards

题目链接：

传送门

题目分析：

线段树妙题，感觉思路奇奇怪怪的，虽然对我来说不是“线段树菜题”（\(ldx\)神仙\(blog\)原话）\(QAQ\)

考虑怎么样维护可合并的信息解决这道题

首先有一个很明显的贪心，一张卡片正反面肯定是能小就小，不带修的话直接就过了

带修的话怎么处理呢，考虑在线段树上维护一个\(sum[0/1]\)，表示这个节点\(l\)位置上卡片选正/反面的时候\(r\)位置上卡片的最小取值，不合法赋\(INF\)

重点在\(pushup\)里面，每次\(pushup\)的时候把左子区间维护的两个\(sum\)分别和右子区间左端点的两个值比较一下看合不合法，不合法赋\(INF\)（感觉说的不是很清楚，自己脑补一下\(or\)看代码可能会好懂一点

每次交换可以看做单点修改

代码：

#include<bits/stdc++.h>
#define N (300000 + 10)
using namespace std;
inline int read() {
	int cnt = 0, f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
	return cnt * f;
}
const int INF = 1000000000 + 7;
int n, m;
int x, y;
int a[N], b[N];
struct node {
	int l, r, sum[2];
	#define l(p) tree[p].l
	#define r(p) tree[p].r
	#define sum0(p) tree[p].sum[0]
	#define sum1(p) tree[p].sum[1]
}tree[N << 2];
void pushup(int p) {
	sum0(p) = sum1(p) = INF;
	if (sum0(p << 1) <= a[l(p << 1 | 1)]) sum0(p) = sum0(p << 1 | 1);
	if (sum0(p << 1) <= b[l(p << 1 | 1)]) sum0(p) = min(sum0(p), sum1(p << 1 | 1));
	if (sum1(p << 1) <= a[l(p << 1 | 1)]) sum1(p) = sum0(p << 1 | 1);
	if (sum1(p << 1) <= b[l(p << 1 | 1)]) sum1(p) = min(sum1(p), sum1(p << 1 | 1));
}
void build(int p, int l, int r) {
	l(p) = l, r(p) = r;
	if (l == r) {sum0(p) = a[l], sum1(p) = b[l]; return;}
	int mid = (l + r) >> 1;
	build (p << 1, l, mid);
	build (p << 1 | 1, mid + 1, r);
	pushup(p);
}
void modify(int p, int x, int u, int v) {
	if (u > v) swap(u, v);
	if (l(p) == r(p)) {sum0(p) = a[l(p)] = u, sum1(p) = b[l(p)] = v; return;}
	int mid = (l(p) + r(p)) >> 1;
	if (x <= mid) modify(p << 1, x, u, v);
	else modify(p << 1 | 1, x, u, v);
	pushup(p);
}
int cura, curb;
int main() {
	n = read();
	for (register int i = 1; i <= n; i++) {
		a[i] = read(), b[i] = read();
		if (a[i] > b[i]) swap(a[i], b[i]);
	}
	build(1, 1, n);
	m = read();
	for (register int i = 1; i <= m; i++) {
		x = read(), y = read();
		cura = a[x], curb = b[x];
		modify(1, x, a[y], b[y]), modify(1, y, cura, curb);
		if (sum0(1) == INF && sum1(1) == INF) printf("NIE\n");
		else printf("TAK\n");
	}
	return 0;
}