UVA - 1400"Ray, Pass me the dishes!"(线段树)
题目链接
题目大意：给你N个数字，要求你动态的给出L到R之间，X>= L && Y<=R,使得X，Y这段的连续和是LR之间的最大连续和，如果有多解，输出X小的，接着是Y小的。
解题思路:结点保存三个附加线段，max_sub, max_suffix, max_prefix.对于每次查询最大的连续和要不出现在左子树的max_sub, 要不就是右子树的max_sub, 要不就是左子树的max_suffix + 右子树的max_prefix.然后每次查询的时候都返回一个新的节点，是新的控制范围和在这个范围内的对应的三个线段值。求max_prefix和max_suffix的时候要注意。
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 5e5 + 5;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1) + 1)
ll A[N], S[N];
struct Segment {
ll v;
int l, r;
Segment(int l = 0, int r = 0, ll v = 0) {
this->l = l;
this->r = r;
this->v = v;
}
Segment operator + (const Segment& a) const{
Segment ans;
ans.l = min (l, a.l);
ans.r = max (r, a.r);
ans.v = v + a.v;
return ans;
}
bool operator < (const Segment &a) const {
if (v == a.v) {
if (l == a.l)
return r > a.r;
return l > a.l;
}
return v < a.v;
}
};
struct Node {
int l, r;
Segment max_sub, max_prefix, max_suffix;
void set (int l, int r, Segment max_sub, Segment max_prefix, Segment max_suffix) {
this->l = l;
this->r = r;
this->max_sub = max_sub;
this->max_prefix = max_prefix;
this->max_suffix = max_suffix;
}
}node[4 * N];
Node Seg_merge(Node a, Node b) {
Node ret;
ll suml = S[a.r] - S[a.l - 1];
ll sumr = S[b.r] - S[b.l - 1];
ret.l = a.l;
ret.r = b.r;
ret.max_sub = max(a.max_suffix + b.max_prefix, max(a.max_sub, b.max_sub));
ret.max_prefix = max (a.max_prefix, Segment(a.l, a.r, suml) + b.max_prefix);
ret.max_suffix = max (b.max_suffix, a.max_suffix + Segment(b.l, b.r, sumr));
return ret;
}
void build (int u, int l, int r) {
if (l == r) {
Segment max_sub(l, r, A[l]);
Segment max_prefix(l, r, A[l]);
Segment max_suffix(l, r, A[l]);
node[u].set(l, r, max_sub, max_prefix, max_suffix);
} else {
int m = (l + r) / 2;
build(lson(u), l, m);
build(rson(u), m + 1, r);
node[u] = Seg_merge(node[lson(u)], node[rson(u)]);
}
}
Node Query (int u, int ql, int qr) {
if (ql <= node[u].l && qr >= node[u].r)
return node[u];
int m = (node[u].l + node[u].r) / 2;
if (ql > m)
return Query (rson(u), ql, qr);
else if (qr <= m)
return Query (lson(u), ql, qr);
else
return Seg_merge(Query(lson(u), ql, qr), Query(rson(u), ql, qr));
}
int n, m;
int main () {
int cas = 0;
int l, r;
Node ans;
while (scanf ("%d%d", &n, &m) != EOF) {
S[0] = 0;
for (int i = 1; i <= n; i++) {
scanf ("%lld", &A[i]);
S[i] = S[i - 1] + A[i];
}
printf ("Case %d:
", ++cas);
build(1, 1, n);
for (int i = 0; i < m; i++) {
scanf ("%d%d", &l, &r);
ans = Query(1, l, r);
printf ("%d %d
", ans.max_sub.l, ans.max_sub.r);
}
}
return 0;
}
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