zoukankan      html  css  js  c++  java
  • SPOJ 1716 Can you answer these queries III

    SPOJ_1716

        这个题目和SPOJ_1043的GSS1是类似的,只不过增加了单点修改的功能。用线段树实现相应的功能即可。

    #include<stdio.h>
    #include<string.h>
    #define MAXD 50010
    #define INF 0x3f3f3f3f3f3f3f3fll
    int N, M, a[MAXD];
    long long lc[4 * MAXD], rc[4 * MAXD], mc[4 * MAXD], sum[4 * MAXD];
    long long Max(long long x, long long y)
    {
        return x > y ? x : y;
    }
    void update(int cur)
    {
        int ls = cur << 1, rs = cur << 1 | 1;
        sum[cur] = sum[ls] + sum[rs];
        mc[cur] = Max(mc[ls], mc[rs]);
        mc[cur] = Max(mc[cur], rc[ls] + lc[rs]);
        lc[cur] = Max(lc[ls], sum[ls] + lc[rs]);
        rc[cur] = Max(rc[rs], sum[rs] + rc[ls]);
    }
    void build(int cur, int x, int y)
    {
        int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
        if(x == y)
        {
            sum[cur] = mc[cur] = lc[cur] = rc[cur] = a[x];
            return ;
        }
        build(ls, x, mid);
        build(rs, mid + 1, y);
        update(cur);
    }
    void refresh(int cur, int x, int y, int k, int v)
    {
        int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
        if(x == y)
        {
            sum[cur] = mc[cur] = lc[cur] = rc[cur] = v;
            return ;
        }
        if(k <= mid)
            refresh(ls, x, mid, k, v);
        else
            refresh(rs, mid + 1, y, k, v);
        update(cur);
    }
    long long Search(int cur, int x, int y, int s, int t, long long &ans, int flag)
    {
        int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
        if(x >= s && y <= t)
        {
            ans = Max(ans, mc[cur]);
            return flag ? rc[cur] : lc[cur];
        }
        if(mid >= t)
            return Search(ls, x, mid, s, t, ans, 0);
        else if(mid + 1 <= s)
            return Search(rs, mid + 1, y, s, t, ans, 1);
        long long ln, rn;
        ln = Search(ls, x, mid ,s, t, ans, 1), rn = Search(rs, mid + 1, y, s, t, ans, 0);
        ans = Max(ans, ln + rn);
        if(flag)
            return Max(sum[rs] + ln, rc[rs]);
        else
            return Max(sum[ls] + rn, lc[ls]);
    }
    void init()
    {
        int i;
        for(i = 1; i <= N; i ++)
            scanf("%d", &a[i]);
        build(1, 1, N);
    }
    void solve()
    {
        int i, k, x, y, q;
        long long ans;
        scanf("%d", &q);
        for(i = 0; i < q; i ++)
        {
            scanf("%d%d%d", &k, &x, &y);
            if(k == 0)
                refresh(1, 1, N, x, y);
            else
            {
                ans = -INF;
                Search(1, 1, N, x, y, ans, 0);
                printf("%lld\n", ans);
            }
        }
    }
    int main()
    {
        while(scanf("%d", &N) == 1)
        {
            init();
            solve();
        }
        return 0;
    }
  • 相关阅读:
    HDU 5744
    HDU 5815
    POJ 1269
    HDU 5742
    HDU 4609
    fzu 1150 Farmer Bill's Problem
    fzu 1002 HangOver
    fzu 1001 Duplicate Pair
    fzu 1150 Farmer Bill's Problem
    fzu 1182 Argus 优先队列
  • 原文地址:https://www.cnblogs.com/staginner/p/2526172.html
Copyright © 2011-2022 走看看