牛客多校第三场 G Removing Stones(分治+线段树)

牛客多校第三场 G Removing Stones(分治+线段树)

题意:

给你n个数,问你有多少个长度不小于2的连续子序列,使得其中最大元素不大于所有元素和的一半

题解:

分治+线段树

线段树维护最大值的位置,每次分治找就找这个最大值,然后看最大值在哪个序列是可行的

怎么看最大值所在的序列是否可行呢?

我们用一个前缀和维护区间和
\[ max<=\frac{1}{2}(sum[r]-sum[l])\\ 2*max-(sum[r]-sum[l])<=0\\ \]
这个最大值在这一段区间内都有可能出现

为了得到总的方案数

我们在当前区间内枚举靠近最大值出现的位置pos的那一段,然后二分右\左端点,即可得到一段最小的区间[i,t]或者[t,i]可以满足 区间内最大元素不大于所有元素和的一半的条件,然后计数就加上容斥后的r-t+1或者t-l+1这一段即可

代码:

/**
 *        ┏┓    ┏┓
 *        ┏┛┗━━━━━━━┛┗━━━┓
 *        ┃       ┃  
 *        ┃   ━    ┃
 *        ┃ >   < ┃
 *        ┃       ┃
 *        ┃... ⌒ ...  ┃
 *        ┃       ┃
 *        ┗━┓   ┏━┛
 *          ┃   ┃ Code is far away from bug with the animal protecting          
 *          ┃   ┃   神兽保佑,代码无bug
 *          ┃   ┃           
 *          ┃   ┃        
 *          ┃   ┃
 *          ┃   ┃           
 *          ┃   ┗━━━┓
 *          ┃       ┣┓
 *          ┃       ┏┛
 *          ┗┓┓┏━┳┓┏┛
 *           ┃┫┫ ┃┫┫
 *           ┗┻┛ ┗┻┛
 */
// warm heart, wagging tail,and a smile just for you!
//
//                            _ooOoo_
//                           o8888888o
//                           88" . "88
//                           (| -_- |)
//                           O\  =  /O
//                        ____/`---'\____
//                      .'  \|     |//  `.
//                     /  \|||  :  |||//  \
//                    /  _||||| -:- |||||-  \
//                    |   | \  -  /// |   |
//                    | \_|  ''\---/''  |   |
//                    \  .-\__  `-`  ___/-. /
//                  ___`. .'  /--.--\  `. . __
//               ."" '<  `.___\_<|>_/___.'  >'"".
//              | | :  `- \`.;`\ _ /`;.`/ - ` : | |
//              \  \ `-.   \_ __\ /__ _/   .-` /  /
//         ======`-.____`-.___\_____/___.-`____.-'======
//                            `=---='
//        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
//                     佛祖保佑      永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
    return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
    return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
    double ans = 1.0;
    while(b) {
        if(b % 2)ans = ans * a;
        a = a * a;
        b /= 2;
    } return ans;
}
LL quick_pow(LL x, LL y) {
    LL ans = 1;
    while(y) {
        if(y & 1) {
            ans = ans * x % mod;
        } x = x * x % mod;
        y >>= 1;
    } return ans;
}
pii Max[maxn << 2];

LL sum[maxn];
int a[maxn];
void push_up(int rt) {
    Max[rt] = max(Max[ls], Max[rs]);
}
void build(int l, int r, int rt) {
    if(l == r) {
        Max[rt] = make_pair(a[l], l);
        return;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    push_up(rt);
}
pii query(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) {
        return Max[rt];
    }
    int mid = (l + r) >> 1;
    pii ans = make_pair(0, 0);
    if(L <= mid) ans = max(ans, query(L, R, lson));
    if(R > mid) ans = max(ans, query(L, R, rson));
    return ans;
}
//二分最短区间大于后缀
int find1(int l, int r, LL x) {
    int L = l, R = r + 1, res = l - 1;
    while (L <= R) {
        int mid = (L + R) >> 1;
        if (sum[r] - sum[mid - 1] >= x) {
            L = mid + 1;
            res = mid;
        } else R = mid - 1;
    }
    return res;
}
int find2(int l, int r, LL x) {
    int L = l - 1, R = r, res = r + 1;
    while (L <= R) {
        int mid = (L + R) >> 1;
        if (sum[mid] - sum[l - 1] >= x) {
            R = mid - 1;
            res = mid;
        } else L = mid + 1;
    }
    return res;
}
LL ans = 0;
int n;
void solve(int l, int r) {
    if(l == r) return;
    int mid = query(l, r, 1, n, 1).second;

    if(l < mid) solve(l, mid - 1);
    if(r > mid) solve(mid + 1, r);
    if(mid - l < r - mid) {
        for(int i = l; i <= mid; i++) {
            int t = find2(mid + 1, r, 2 * a[mid] - (sum[mid] - sum[i - 1]));
            ans += r - t + 1;
        }
    } else {
        for(int i = mid; i <= r; i++) {
            int t = find1(l, mid - 1, 2 * a[mid] - (sum[i] - sum[mid - 1]));
            ans += t - l + 1;
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        sum[0] = 0;
        ans = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            sum[i] = sum[i - 1] + a[i];
        }
        build(1, n, 1);
        solve(1, n);
        printf("%lld\n", ans);
    }
    return 0;
}
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