题目链接
题解
思来想去，发现很显然可以用线段树维护
每个区间保存所有合法方案的左右端点【当左端点一定是，右端点当然存最小的那个就行了】
这么整的数，(frac{1}{1000})的概率啊23333
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 200005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int ln[maxn << 2][2],rn[maxn << 2][2],cnt[maxn << 2];
int a[maxn],b[maxn],n,m;
void upd(int u){
if (!cnt[ls] || !cnt[rs]) cnt[u] = 0;
else {
cnt[u] = 0;
for (int i = 0; i < cnt[ls]; i++){
int r = INF;
for (int j = 0; j < cnt[rs]; j++){
if (rn[ls][i] <= ln[rs][j])
r = min(r,rn[rs][j]);
}
if (r < INF){
ln[u][cnt[u]] = ln[ls][i];
rn[u][cnt[u]] = r;
cnt[u]++;
}
}
}
}
void build(int u,int l,int r){
if (l == r){
cnt[u] = 2;
ln[u][0] = rn[u][0] = a[l];
ln[u][1] = rn[u][1] = b[l];
return;
}
int mid = l + r >> 1;
build(ls,l,mid);
build(rs,mid + 1,r);
upd(u);
}
void modify(int u,int l,int r,int pos,cp v){
if (l == r){
cnt[u] = 2;
ln[u][0] = rn[u][0] = v.first;
ln[u][1] = rn[u][1] = v.second;
return;
}
int mid = l + r >> 1;
if (mid >= pos) modify(ls,l,mid,pos,v);
else modify(rs,mid + 1,r,pos,v);
upd(u);
}
void print(int u,int l,int r){
if (l == r){
printf("(%d,%d) ",ln[u][0],ln[u][1]);
return;
}
int mid = l + r >> 1;
print(ls,l,mid); print(rs,mid + 1,r);
}
int main(){
n = read();
REP(i,n) a[i] = read(),b[i] = read();
build(1,1,n);
m = read();
int c,d;
while (m--){
//print(1,1,n); puts("");
c = read(); d = read();
modify(1,1,n,c,mp(a[d],b[d]));
modify(1,1,n,d,mp(a[c],b[c]));
swap(a[c],a[d]); swap(b[c],b[d]);
puts(cnt[1] ? "TAK" : "NIE");
}
return 0;
}