Gym 101630G The Great Wall

题意:

三个有n个正整数的数组满足a[i]

分析:

首先把b[i],c[i]减去a[i],最后再加回来方便计算

这种第k大的经典套路就是二分答案check有多少个值比他小

在这个题的check里,我们分两部分讨论:x和y的区间有重叠的和没有重叠的,对于没有重叠的十分容易,枚举每个点作为右区间的起始端点再离散化区间b[i]的和用树状数组统计与多少个前面的区间值相加

对于重叠的,表面上看重叠区间之间会互相影响,无法像不重叠那样直接将区间b[i]和插入树状数组

这就需要将加入树状数组的值凑成一个只与当前点相关的值,具体凑法如下:

 

#include
#include
#include
#include
#include
#define sc(x) scanf("%d", &x)
#define pb push_back
#define ALL(x) x.begin(),x.end()
using namespace std;

typedef long long LL;
const int maxn = 3e4+10;
int n, r, a[maxn], b[maxn], c[maxn];
LL bit[maxn*3], k, preb[maxn], prec[maxn];
vector val;

inline int lowbit(int x){return x&-x;}
void add(int x, int v){
    for(int i = x; i < maxn*3; i += lowbit(i))
        bit[i] += v;
}
LL sum(int x){
    LL res = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        res += bit[i];
    return res;
}

LL check(LL mid){
    //printf("check(%lld):",mid);
    memset(bit, 0, sizeof(bit));
    LL res = 0;
    for(int i = r+1, j = 1; i+r-1 <= n; i++, j++){
        int id = lower_bound(ALL(val),preb[j+r-1]-preb[j-1])-val.begin()+1;
        add(id,1);
        LL tmp = preb[i+r-1]-preb[i-1];
        tmp = mid-tmp;
        id = lower_bound(ALL(val),tmp)-val.begin();
        res += sum(id);
    }
    //printf("%lld,", res);
    memset(bit, 0, sizeof(bit));
    for(int i = 2; i+r-1 <= n; i++){
        LL tmp = prec[i+r-2] - preb[i+r-2] - preb[i-2];
        int id = lower_bound(ALL(val),tmp)-val.begin()+1;
        add(id,1);
        if(i-r >= 1){
            tmp = prec[i-r+r-1] - preb[i-r+r-1] - preb[i-r-1];
            id = lower_bound(ALL(val),tmp)-val.begin()+1;
            add(id, -1);
        }
        tmp = preb[i+r-1] + preb[i-1] - prec[i-1];
        tmp = mid-tmp;
        id = lower_bound(ALL(val),tmp)-val.begin();
        res += sum(id);
    }
    //printf("%lld\n", res);
    return res;
}

int main(){
    sc(n); sc(r); cin >> k; k--;
    LL sum = 0;
    for(int i = 1; i <= n; i++) sc(a[i]), sum += a[i];
    for(int i = 1; i <= n; i++) sc(b[i]), b[i] -= a[i];
    for(int i = 1; i <= n; i++) sc(c[i]), c[i] -= a[i];
    for(int i = 1; i <= n; i++){
        preb[i] = preb[i-1] + b[i];
        prec[i] = prec[i-1] + c[i];
        if(i >= r){
            val.pb(prec[i]-preb[i-r]-preb[i]);
    		val.pb(preb[i]-preb[i-r]);
        }
    }
    sort(ALL(val));
    val.erase(unique(ALL(val)), val.end());
    LL L = 0, R = 3e10+10;
    while(L+1 < R){
        LL mid = (L+R)/2;
        //printf("check(%lld):%lld, k:%lld\n",mid,check(mid),k);
        if(check(mid) > k) R = mid;
        else L = mid;
    }
    cout << L+sum << endl;
    return 0;
}