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  • 借教室(线段树)

    题目链接:https://ac.nowcoder.com/acm/problem/16564

    题目大意: 有n天可以租教室,给出每天可以租用的教室数量,有m次询问,找出最小值判断能不能完成租借。
    思路:线段树求区间最小值。(模板)

    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <string>
    #include <stack>
    #include <queue>
    #include <cmath>
    #define ll long long
    #define pi 3.1415927
    #define inf 0x3f3f3f3f
    #define mod 1000000007
    using namespace std;
    int n,m;
    int a[1000005];
    struct node{
        int l,r,min,f;
    }tree[4000005];
    void build(int i, int l, int r)  //建树
    {
        tree[i].l=l,tree[i].r=r;
        if(l==r)
        {
            tree[i].min=a[l];
            return ;
        }
        int mid=(l+r)>>1;
        build(i<<1,l,mid);
        build(i<<1|1,mid+1,r);
        tree[i].min=min(tree[i<<1].min,tree[i<<1|1].min);
    }
    void down(int i) //懒节点下穿
    {
        tree[i<<1].f+=tree[i].f;
        tree[i<<1|1].f+=tree[i].f;
        tree[i<<1].min-=tree[i].f;
        tree[i<<1|1].min-=tree[i].f;
        tree[i].f=0;
    }
    void add(int p, int j, int k, int i) //维护区间
    {
        if(tree[i].l>=j &&tree[i].r<=k)
        {
            tree[i].min-=p;
            tree[i].f+=p;
            return ;
        }
        if(tree[i].f)
            down(i);
        if(tree[i<<1].r>=j)
            add(p,j,k,i<<1);
        if(tree[i<<1|1].l<=k)
            add(p,j,k,i<<1|1);
        tree[i].min=min(tree[i<<1].min,tree[i<<1|1].min);
    }
    int mins=inf;
    void ask(int j, int k, int i) //找区间最小值
    {
        if(tree[i].l>=j && tree[i].r<=k)
        {
            mins=min(mins,tree[i].min);
            return ;
        }
        if(tree[i].f)
            down(i);
        if(tree[i<<1].r>=j)
            ask(j,k,i<<1);
        if(tree[i<<1|1].l<=k)
            ask(j,k,i<<1|1);
    }
    int main ()
    {
        int T,i,t,j,k,p,sum=0;
        cin>>n>>m;
        for(i=1;i<=n;++i)
            scanf("%d",&a[i]);
        build(1,1,n);
        for(i=0;i<m;++i)
        {
            scanf("%d %d %d",&p,&j,&k); 
            mins=inf;
            ask(j,k,1);
            if(mins<p){
                cout<<"-1"<<endl<<i+1<<endl;
                return 0;
            }
            add(p,j,k,1);
        }
        cout<<"0"<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/blowhail/p/13264256.html
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