• Luogu3527:[POI2011]MET-Meteors


    题面

    Luogu

    Sol

    整体二分
    比较简单,当练手题
    每次树状数组统计

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(3e5 + 5); 
    
    IL ll Input(){
        RG ll x = 0, z = 1; RG char c = getchar();
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
    
    int n, m, k, ql[_], qr[_], qv[_], ans[_];
    vector <int> son[_];
    struct Data{
        int id;
        ll v;
    } p[_], q1[_], q2[_];
    ll bit[_];
    
    IL void Modify(RG int x, RG ll v){
        for(; x <= m; x += x & -x) bit[x] += v;
    }
    
    IL ll Query(RG int x){
        RG ll ret = 0;
        for(; x; x -= x & -x) ret += bit[x];
        return ret;
    }
    
    IL void Solve(RG int l, RG int r, RG int L, RG int R){
        if(L > R) return;
        if(l == r){
            for(RG int i = L; i <= R; ++i) ans[p[i].id] = l;
            return;
        }
        RG int mid = (l + r) >> 1, t1 = 0, t2 = 0;
        for(RG int i = l; i <= mid; ++i)
            if(ql[i] <= qr[i])
                Modify(ql[i], qv[i]), Modify(qr[i] + 1, -qv[i]);
            else Modify(1, qv[i]), Modify(qr[i] + 1, -qv[i]), Modify(ql[i], qv[i]);
        for(RG int i = L; i <= R; ++i){
            RG ll sum = 0;
            for(RG int j = 0, len = son[p[i].id].size(); j < len; ++j)
                sum += Query(son[p[i].id][j]);
            if(p[i].v <= sum) q1[++t1] = p[i];
            else p[i].v -= sum, q2[++t2] = p[i];
        }
        for(RG int i = l; i <= mid; ++i)
            if(ql[i] <= qr[i])
                Modify(ql[i], -qv[i]), Modify(qr[i] + 1, qv[i]);
            else Modify(1, -qv[i]), Modify(qr[i] + 1, qv[i]), Modify(ql[i], -qv[i]);
        for(RG int i = L, j = 1; j <= t1; ++i, ++j) p[i] = q1[j];
        for(RG int i = L + t1, j = 1; j <= t2; ++i, ++j) p[i] = q2[j];
        Solve(l, mid, L, L + t1 - 1); Solve(mid + 1, r, L + t1, R);
    }
    
    int main(RG int argc, RG char* argv[]){
        n = Input(); m = Input();
        for(RG int i = 1; i <= m; ++i) son[Input()].push_back(i);
        for(RG int i = 1; i <= n; ++i) p[i].id = i, p[i].v = Input();
        k = Input();
        for(RG int i = 1; i <= k; ++i)
            ql[i] = Input(), qr[i] = Input(), qv[i] = Input();
        ql[k + 1] = 1; qr[k + 1] = m; qv[k + 1] = 1e9;
        Solve(1, k + 1, 1, n);
        for(RG int i = 1; i <= n; ++i)
            if(ans[i] == k + 1) puts("NIE");
            else printf("%d
    ", ans[i]);
        return 0;
    }
    
  • 相关阅读:
    土木工程材料0732
    07 具有无关项的逻辑函数及其化简
    06 逻辑函数化简法
    AD中板子挖孔开槽
    电容式触摸按键原理
    LTspice
    三相电
    7、简单电阻容元件模型的创建
    cadence17.4在笔记本下设置菜单显示不全的解决办法
    36. 二叉搜索树与双向链表
  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8423554.html
Copyright © 2020-2023  润新知