• HDU 4027 Can you answer these queries? 线段树


    直接暴力更新,复杂度10NlogN

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <map>
    #include <set>
    #include <vector>
    #include <string>
    #include <queue>
    #include <deque>
    #include <bitset>
    #include <list>
    #include <cstdlib>
    #include <climits>
    #include <cmath>
    #include <ctime>
    #include <algorithm>
    #include <stack>
    #include <sstream>
    #include <numeric>
    #include <fstream>
    #include <functional>
    
    using namespace std;
    
    #define MP make_pair
    #define PB push_back
    #define lson rt << 1, l, mid
    #define rson rt << 1 | 1, mid + 1, r
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector<int> VI;
    typedef pair<int,int> pii;
    const int INF = INT_MAX / 3;
    const double eps = 1e-8;
    const LL LINF = 1e17;
    const double DINF = 1e60;
    const int maxn = 1e5 + 10;
    const int maxP = 11;
    
    int n;
    LL sum[maxn << 2],maxv[maxn << 2],val[maxn << 2];
    
    void pushup(int rt) {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
        maxv[rt] = max(maxv[rt << 1],maxv[rt << 1 | 1]);
    }
    
    void build(int rt,int l,int r) {
        int mid = (l + r) >> 1;
        if(l == r) {
            sum[rt] = maxv[rt] = val[l];
        }
        else {
            build(lson); build(rson);
            pushup(rt);
        }
    }
    
    void update(int rt,int l,int r,int ql,int qr) {
        if(l == r || (ql <= l && qr >= r && maxv[rt] == 1)) {
            if(maxv[rt] == 1) return;
            sum[rt] = sqrt((double)sum[rt]);
            maxv[rt] = sum[rt];
        }
        else {
            int mid = (l + r) >> 1;
            if(ql <= mid) update(lson,ql,qr);
            if(qr > mid) update(rson,ql,qr);
            pushup(rt);
        }
    }
    
    LL query(int rt,int l,int r,int ql,int qr) {
        if(ql <= l && qr >= r) return sum[rt];
        int mid = (l + r) >> 1;
        LL ret = 0;
        if(ql <= mid) ret += query(lson,ql,qr);
        if(qr > mid) ret += query(rson,ql,qr);
        return ret;
    }
    
    
    int main() {
    //    freopen("/tmp/input","r",stdin);
    //    freopen("/tmp/output","w",stdout);
        int kase = 1;
        while(scanf("%d",&n) != EOF) {
            cout << "Case #" << kase++ << ":" << endl;
            for(int i = 1;i <= n;i++) {
                //cin >> val[i];
                scanf("%I64d",&val[i]);
            }
            build(1,1,n);
            int cmd,l,r,Q; cin >> Q;
            for(int i = 0;i < Q;i++) {
                //cin >> cmd >> l >> r;
                scanf("%d%d%d",&cmd,&l,&r);
                if(l > r) swap(l,r);
                if(cmd == 0) update(1,1,n,l,r);
                //else cout << query(1,1,n,l,r) << endl;
                else printf("%I64d
    ",query(1,1,n,l,r));
            }
            printf("
    ");
        }
        return 0;
    }
    

      

  • 相关阅读:
    linux Segmentation faults 段错误详解
    linux cut
    linux sed
    linux tr
    linux ar
    objdump--反汇编查看
    linux中dd命令
    readelf
    登录后,前端做了哪些工作,如何得知已登录?
    正向代理和反向代理?
  • 原文地址:https://www.cnblogs.com/rolight/p/3927047.html
Copyright © 2020-2023  润新知