Ultraviolet A 1400, uva1400

Source: Internet
Author: User

Ultraviolet A 1400, uva1400
UV 1400-"Ray, Pass me the dishes! "

Question Link

Question: Given a sequence, ask for a [L, R] interval each time, and find the maximum continuous subsequence and

Idea: Line Segment tree, each node maintains three values, the maximum continuous subsequence, the maximum continuous prefix sequence, and the maximum continuous suffix sequence. When pushup is performed, create a new node based on the three sequences.


#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson(x) ((x<<1) + 1)#define rson(x) ((x<<1) + 2)#define MP(a, b) make_pair(a, b)typedef long long ll;typedef pair<int, int> Point;const int N = 500005;int n, m;ll a[N], sum[N];struct Node {    int l, r;    int prex, sufx;    Point sub;} node[4 * N];ll get(Point x) {    return sum[x.second] - sum[x.first - 1];}bool Max(Point a, Point b) {    long long sa = get(a);    long long sb = get(b);    if (sa != sb) return sa > sb;    return a < b;}Point Maxsub(Node a, Node b) {    Point ans;    if (Max(a.sub, b.sub)) ans = a.sub;    else ans = b.sub;    if (Max(MP(a.sufx, b.prex), ans)) ans = MP(a.sufx, b.prex);    return ans;}int Maxpre(Node a, Node b) {    Point ans = MP(a.l, a.prex);    if (Max(MP(a.l, b.prex), ans)) ans = MP(a.l, b.prex);    return ans.second;}int Maxsuf(Node a, Node b) {    Point ans = MP(b.sufx, b.r);    if (Max(MP(a.sufx, b.r), ans)) ans = MP(a.sufx, b.r);    return ans.first;}Node pushup(Node a, Node b) {    Node ans;    ans.l = a.l; ans.r = b.r;    ans.sub = Maxsub(a, b);    ans.prex = Maxpre(a, b);    ans.sufx = Maxsuf(a, b);    return ans;}void build(int l, int r, int x) {    if (l == r) {node[x].l = l; node[x].r = r;node[x].prex = node[x].sufx = l;node[x].sub = MP(l, l);return ;    }    int mid = (l + r) / 2;    build(l, mid, lson(x));    build(mid + 1, r, rson(x));    node[x] = pushup(node[lson(x)], node[rson(x)]);}Node Query(int l, int r, int x) {    if (l <= node[x].l && r >= node[x].r)return node[x];    int mid = (node[x].l + node[x].r) / 2;    Node ans;    if (l <= mid && r > mid)ans = pushup(Query(l, r, lson(x)), Query(l, r, rson(x)));    else if (l <= mid) ans = Query(l, r, lson(x));    else if (r > mid) ans = Query(l, r, rson(x));    return ans;}int main() {    int cas = 0;    while (~scanf("%d%d", &n, &m)) {for (int i = 1; i <= n; i++) {    scanf("%lld", &a[i]);    sum[i] = sum[i - 1] + a[i];}build(1, n, 0);printf("Case %d:\n", ++cas);int a, b;while (m--) {    scanf("%d%d", &a, &b);    Node ans = Query(a, b, 0);    printf("%d %d\n", ans.sub.first, ans.sub.second);}    }    return 0;}

1400*1400 pixels or more JPG Images

Choose among them, which is definitely eye-catching, and can be used as a cover on a wide range of topics.

Beautiful but green plants-based scenery may have a dull cover, unless you find the best

A vertical chart can be used as long as the title or work name is associated with the eyes. It is not difficult for the eyes or observations to relate to various things.

Square chart

A thoughtful subject can be configured with a cover

Travel, stories, etc.

1400/40 Simple Calculation Method

1400/40 = (1200 + 200)/40 = 1200/40 + 200/40 = 30 + 5 = 35

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