• 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
    ",ans);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/9237424.html
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