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  • BZOJ5343 [Ctsc2018]混合果汁 【二分 + 主席树】

    题目链接

    BZOJ5343

    题解

    明显要二分一下美味度,然后用尽量少的价格去购买饮料,看看能否买到(L)升,然后看看能否控制价格在(g)
    尽量少的价格,就优先先选完便宜的饮料,由于询问的是一定美味度范围的,主席树上询问即可

    #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
    using namespace std;
    const int maxn = 100005,maxm = 5000005;
    const LL INF = 1000000000000000001ll;
    inline LL read(){
    	LL 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 n,m;
    struct node{
    	int d,p,l;
    }e[maxn];
    inline bool operator <(const node& a,const node& b){
    	return a.d < b.d;
    }
    int Maxp;
    LL S[maxn],sum[maxm],val[maxm];
    int ls[maxm],rs[maxm],cnt,rt[maxn];
    void add(int& u,int pre,int l,int r,int pos,int v){
    	sum[u = ++cnt] = sum[pre]; val[u] = val[pre];
    	ls[u] = ls[pre]; rs[u] = rs[pre];
    	val[u] += 1ll * pos * v; sum[u] += v;
    	if (l == r) return;
    	int mid = l + r >> 1;
    	if (mid >= pos) add(ls[u],ls[pre],l,mid,pos,v);
    	else add(rs[u],rs[pre],mid + 1,r,pos,v);
    }
    LL query(int u,int v,int l,int r,LL ml){
    	if (!u) return 0;
    	if (l == r) return ml * l;
    	int mid = l + r >> 1;
    	LL t = sum[ls[u]] - sum[ls[v]];
    	if (t >= ml) return query(ls[u],ls[v],l,mid,ml);
    	return val[ls[u]] - val[ls[v]] + query(rs[u],rs[v],mid + 1,r,ml - t);
    }
    bool check(int d,LL g,LL L){
    	int pre = lower_bound(e + 1,e + 1 + n,(node){d,0,0}) - e - 1;
    	if (S[n] - S[pre] < L) return false;
    	return query(rt[n],rt[pre],1,Maxp,L) <= g;
    }
    int main(){
    	n = read(); m = read();
    	int M = 0;
    	REP(i,n){
    		e[i].d = read();
    		e[i].p = read();
    		e[i].l = read();
    		M = max(M,e[i].d);
    		Maxp = max(Maxp,e[i].p);
    	}
    	sort(e + 1,e + 1 + n);
    	for (int i = 1; i <= n; i++)
    		S[i] = S[i - 1] + e[i].l;
    	for (int i = 1; i <= n; i++){
    		add(rt[i],rt[i - 1],1,Maxp,e[i].p,e[i].l);
    	}
    	LL g,L;
    	while (m--){
    		g = read(); L = read();
    		int l = 1,r = M,mid;
    		while (l < r){
    			mid = l + r + 1 >> 1;
    			if (check(mid,g,L)) l = mid;
    			else r = mid - 1;
    		}
    		if (check(l,g,L)) printf("%d
    ",l);
    		else puts("-1");
    	}
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Mychael/p/9068184.html
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