POJ 4047 Garden 线段树

来源：互联网发布：java 量化交易编辑：程序博客网时间：2024/05/07 08:51

这题只要想出怎么建树就简单了吧。

他问的是连续k个值最大的是多少

就可以用1~k之和，2~k+1之和，3~k+2之和.......做为结点。

然后就发现，变成了简单的区间更新和区间查询问题了。

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-5#define MAXN 222222#define MAXM 22222#define INF 1000000007#define lch(x) x<<1#define rch(x) x<<1|1#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;int val[MAXN];int n, k, m;int t[MAXN];int mx[4 * MAXN], cover[4 * MAXN];void up(int rt){    mx[rt] = max(mx[lch(rt)], mx[rch(rt)]);}void down(int rt){    if(cover[rt])    {        cover[lch(rt)] += cover[rt];        cover[rch(rt)] += cover[rt];        mx[lch(rt)] += cover[rt];        mx[rch(rt)] += cover[rt];        cover[rt] = 0;    }}void build(int l, int r, int rt){    cover[rt] = 0;    if(l == r)    {        mx[rt] = t[l];        return;    }    int m = (l + r) >> 1;    build(lson);    build(rson);    up(rt);}void update(int L, int R, int l, int r, int rt, int v){    if(L <= l && R >= r)    {        mx[rt] += v;        cover[rt] += v;        return;    }    down(rt);    int m = (l + r) >> 1;    if(m >= L) update(L, R, lson, v);    if(m < R) update(L, R, rson, v);    up(rt);}int query(int L, int R, int l, int r, int rt){    if(L <= l && R >= r) return mx[rt];    down(rt);    int tmp = -INF;    int m = (l + r) >> 1;    if(m >= L) tmp = max(tmp, query(L, R, lson));    if(m < R) tmp = max(tmp, query(L, R, rson));    return tmp;}void change(int x, int y){    int st = x - k + 1, ed = x;    if(st < 1) st = 1;    update(st, ed, 1, n, 1, y - val[x]);    val[x] = y;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        scanf("%d%d%d", &n, &m, &k);        for(int i = 1; i <= n; i++) scanf("%d", &val[i]);        t[1] = 0;        for(int i = 1; i <= k; i++) t[1] += val[i];        for(int i = 2; i <= n - k + 1; i++) t[i] = t[i - 1] - val[i - 1] + val[i - 1 + k];        n = n - k + 1;        build(1, n, 1);        int op, x, y;        while(m--)        {            scanf("%d%d%d", &op, &x, &y);            if(op == 0) change(x, y);            else if(op == 1)            {                int tmp = val[x];                change(x, val[y]);                change(y, tmp);            }            else if(op == 2) printf("%d\n", query(x, y - k + 1, 1, n, 1));        }    }    return 0;}