UVA-1400 Ray, pass me the dishes, la 3938, line segment tree, interval Query

Question: Give the maximum continuous range of a column (N) and M query interval [L, R] [x, y] (L <= x <= Y <= r ). (N, m <= 500 000)

Idea: The maximum continuous interval of the dynamic query interval;

For the maximum continuous interval and:

Use the line segment tree to maintain the maximum continuity and sum_sub, the maximum prefix and sum_prefix, the maximum suffix, and sum_suffix.

Root. sum_sub = max {L. sum_sub, R. sum_sub, (L. sum_suffix + R. sum_prefix )};

The range of question requirements, similar:

Use the line segment tree to maintain max_sub in the maximum continuous interval, max_prefix at the right endpoint, and max_suffix at the left endpoint with the maximum suffix.

For detailed operations, see the code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define lc rt<<1#define rc rt<<1|1const int maxn = 500000 + 5;typedef long long LL;LL num[maxn], max_prefix[maxn<<2], max_suffix[maxn<<2];struct node{    int l, r;    node(int ll=0, int rr=0):l(ll),r(rr){}} max_sub[maxn<<2];LL prefix_sum[maxn];int  ql, qr;LL sum(int l, int r){    return prefix_sum[r] - prefix_sum[l-1];}LL sum(node a){    return sum(a.l, a.r);}node better(node a, node b){    if(sum(a) != sum(b)) return sum(a) > sum(b) ? a:b;    return (a.l<b.l||(a.l==b.l&&a.r<b.r))? a:b;}void build(int rt, int l, int r){    if(l==r){        max_prefix[rt] = max_suffix[rt] = l;        max_sub[rt] = node(l,l);        return ;    }    int m = (l+r)>>1;    build(lc, l, m);    build(rc, m+1, r);    LL v1 = sum(l, max_prefix[lc]);    LL v2 = sum(l, max_prefix[rc]);    if(v1 == v2) max_prefix[rt] = min(max_prefix[lc], max_prefix[rc]);    else max_prefix[rt] = v1 > v2 ? max_prefix[lc] : max_prefix[rc];    v1 = sum(max_suffix[lc], r);    v2 = sum(max_suffix[rc], r);    if(v1 == v2) max_suffix[rt] = min(max_suffix[lc], max_suffix[rc]);    else max_suffix[rt] = v1 > v2 ? max_suffix[lc] : max_suffix[rc];    max_sub[rt] = better(max_sub[lc], max_sub[rc]);    max_sub[rt] = better(max_sub[rt], node(max_suffix[lc], max_prefix[rc]));}int query_prefix(int rt, int l, int r){    if(qr >= max_prefix[rt]) return max_prefix[rt];    int m = (l+r)>>1;    //l<=qr<=m    if(qr <= m) return query_prefix(lc, l, m);    //m+1<=qr<=r    int rr = query_prefix(rc, m+1, r);    node ret = better(node(l,rr), node(l, max_prefix[lc]));    return ret.r;}int query_suffix(int rt, int l, int r){    if(ql <= max_suffix[rt]) return max_suffix[rt];    int m = (l+r)>>1;    //m+1<=ql<=r    if(ql > m) return query_suffix(rc, m+1, r);    //l<=ql<=m    int ll = query_suffix(lc, l, m);    node ret = better(node(ll, r), node(max_suffix[rc], r));    return ret.l;}node query(int rt, int l, int r){    if(ql <= l && r <= qr) return max_sub[rt];    int m = (l+r)>>1;    if(qr <= m) return query(lc, l, m);    if(ql >  m) return query(rc, m+1, r);    //ql <= m <= qr    int ll = query_suffix(lc, l, m);  //l_max_suffix    int rr = query_prefix(rc, m+1, r); //r_max_prefix    node mid = node(ll, rr);    node sub = better( query(lc, l, m), query(rc, m+1, r));    return better( mid, sub);}int main(){    int n, m, i, cas = 1, l, r;    while(~scanf("%d%d", &n, &m)){        printf("Case %d:\n", cas++);        prefix_sum[0] = 0;        for(i=1; i<=n; ++i) {                scanf("%lld", &num[i]);                prefix_sum[i] = prefix_sum[i-1] + num[i];        }        node v = node(1,3);        build(1, 1, n);        while(m--){            scanf("%d%d", &l, &r);            ql = l; qr = r;            node ans = query(1, 1, n);            printf("%d %d\n", ans.l, ans.r);        }    }    return 0;}