BZOJ1113 [Poi2008]海报PLA 【分治 + 线段树】

题目链接

BZOJ1113

题解

显然只与高有关，每次选择所有海报中最低的覆盖所有海报，然后分治两边

每个位置会被调用一次，复杂度\(O(nlogn)\)

\(upd:\)智障了，，是一道\(O(n)\)普及-贪心模拟题QAQ

#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 = 250005,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 h[maxn],n,ans;
int pos[maxn << 2],mn[maxn << 2],tag[maxn << 2];
inline void upd(int u){
	if (mn[ls] <= mn[rs]) pos[u] = pos[ls];
	else pos[u] = pos[rs];
	mn[u] = min(mn[ls],mn[rs]);
}
inline void pd(int u){
	if (tag[u]){
		mn[ls] += tag[u]; mn[rs] += tag[u];
		tag[ls] += tag[u]; tag[rs] += tag[u];
		tag[u] = 0;
	}
}
void add(int u,int l,int r,int L,int R,int v){
	if (l >= L && r <= R){mn[u] += v; tag[u] += v; return;}
	pd(u);
	int mid = l + r >> 1;
	if (mid >= L) add(ls,l,mid,L,R,v);
	if (mid < R) add(rs,mid + 1,r,L,R,v);
	upd(u);
}
cp query(int u,int l,int r,int L,int R){
	if (l >= L && r <= R) return mp(mn[u],pos[u]);
	pd(u);
	int mid = l + r >> 1;
	if (mid >= R) return query(ls,l,mid,L,R);
	if (mid < L) return query(rs,mid + 1,r,L,R);
	cp t1 = query(ls,l,mid,L,R),t2 = query(rs,mid + 1,r,L,R);
	return mp(min(t1.first,t2.first),t1.first <= t2.first ? t1.second : t2.second);
}
void build(int u,int l,int r){
	if (l == r){mn[u] = h[l]; pos[u] = l; return;}
	int mid = l + r >> 1;
	build(ls,l,mid);
	build(rs,mid + 1,r);
	upd(u);
}
void solve(int l,int r){
	if (l > r) return;
	cp t = query(1,1,n,l,r);
	if (t.first){
		if (t.second > l) add(1,1,n,l,t.second - 1,-t.first);
		if (t.second < r) add(1,1,n,t.second + 1,r,-t.first);
		ans++;
	}
	solve(l,t.second - 1);
	solve(t.second + 1,r);
}
int main(){
	n = read();
	REP(i,n) read(),h[i] = read();
	build(1,1,n);
	solve(1,n);
	printf("%d\n",ans);
	return 0;
}