zoukankan      html  css  js  c++  java
  • Codeforces 1179C Serge and Dining Room 线段树

    Serge and Dining Room

    改变选的顺序, 最后的结果不变。

    这种题一般都和前缀和有关, 建个线段树维护前缀和就好了。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 1e6 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 7340033;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}
    
    int n, m, a[N], b[N];
    
    const int maxVal = 1000000;
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    struct segmentTree {
        int lazy[N << 2], mx[N << 2];
        inline void pull(int rt) {
            mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
        }
        inline void push(int rt) {
            if(lazy[rt]) {
                lazy[rt << 1] += lazy[rt];
                lazy[rt << 1 | 1] += lazy[rt];
                mx[rt << 1] += lazy[rt];
                mx[rt << 1 | 1] += lazy[rt];
                lazy[rt] = 0;
            }
        }
        void update(int L, int R, int val, int l = 1, int r = maxVal, int rt = 1) {
            if(R < l || r < L || R < L) return;
            if(L <= l && r <= R) {
                mx[rt] += val;
                lazy[rt] += val;
                return;
            }
            push(rt);
            int mid = l + r >> 1;
            update(L, R, val, lson);
            update(L, R, val, rson);
            pull(rt);
        }
        int query(int l = 1, int r = maxVal, int rt = 1) {
            if(mx[rt] <= 0) return -1;
            if(l == r) return l;
            push(rt);
            int mid = l + r >> 1;
            if(mx[rt << 1 | 1] > 0) return query(rson);
            else return query(lson);
        }
    } Tree;
    
    
    int main() {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= m; i++) scanf("%d", &b[i]);
        for(int i = 1; i <= n; i++) {
            Tree.update(1, a[i], 1);
        }
        for(int i = 1; i <= m; i++) {
            Tree.update(1, b[i], -1);
        }
        int q; scanf("%d", &q);
        while(q--) {
            int op, p, x;
            scanf("%d%d%d", &op, &p, &x);
            if(op == 1) {
                Tree.update(1, a[p], -1);
                a[p] = x;
                Tree.update(1, a[p], 1);
            } else {
                Tree.update(1, b[p], 1);
                b[p] = x;
                Tree.update(1, b[p], -1);
            }
            printf("%d
    ", Tree.query());
        }
        return 0;
    }
    
    /*
    */
  • 相关阅读:
    spark streaming 入门例子
    ElasticSearch-hadoop saveToEs源码分析
    spark 资源参数调优
    spark 任务运行原理
    spark RDD底层原理
    用实例说明Spark stage划分原理
    Spark任务提交底层原理
    spark shuffle内在原理说明
    iOS 辛格尔顿
    CodeForces 22D Segments 排序水问题
  • 原文地址:https://www.cnblogs.com/CJLHY/p/11102793.html
Copyright © 2011-2022 走看看