BZOJ3526 [Poi2014]Card 【线段树】

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;

}

巴特西

BZOJ3526 [Poi2014]Card 【线段树】

题目链接

题解

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