ACM/ICPC 2018亚洲区预选赛北京赛站网络赛 D - 80 Days

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要求的点满足特定性质
对a[i]-b[i]做前缀和s[i]
满足从该点开始往后n个点的最小s不会比起点的s小超过c
只需满足这个条件就付的起钱

#include<bits/stdc++.h> 
using namespace std;
#define For(i,a,b) for (int i=(a);i<=(b);i++)
#define cint const int &
#define inf 100000000000000000LL
#define N 1000005

typedef long long ll;
ll s[N<<1],ans;
int a[N],b[N],tot,rt,L,R,n,c,m;

struct node{ll k;int l,r;}t[N<<2];

void build(int &k,cint l,cint r)
{
    if (!k) t[k=++tot]=(node){0,0,0};
    if (l==r) {t[k].k=s[l];return;}
    int mid=l+r>>1;
    build(t[k].l,l,mid);
    build(t[k].r,mid+1,r);
    t[k].k=min(t[t[k].l].k,t[t[k].r].k);
}

inline void prepare()
{
    tot=0;
    rt=0;
}

void find(cint k,cint l,cint r)
{
    if (L<=l && r<=R)
    {
        ans=min(ans,t[k].k);
        return;
    }
    int mid=l+r>>1;
    if (L<=mid) find(t[k].l,l,mid);
    if (mid<R) find(t[k].r,mid+1,r);
}

inline bool check(int st)
{
    L=st;
    R=st+n-1;
    ans=inf;
    find(rt,1,m);
    return ans+c>=s[st-1];
}

inline void solve()
{
    prepare();
    scanf("%d%d",&n,&c);
    For(i,1,n) scanf("%d",a+i);
    For(i,1,n) scanf("%d",b+i);
    For(i,1,n) s[i]=s[n+i]=a[i]-b[i];
    m=n<<1;
    For(i,1,m) s[i]+=s[i-1];
    build(rt,1,m);
    For(i,1,n) if (check(i)) {printf("%d\n",i);return;}
    puts("-1");
}

int main()
{
    int T;
    scanf("%d",&T);
    while (T--) solve();
}

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转载自blog.csdn.net/Pure_W/article/details/83552479