UVA 1400 - "Ray, Pass me the dishes!"
题意:给定一个序列,每次询问一个[L,R]区间。求出这个区间的最大连续子序列和
思路:线段树,每一个节点维护3个值。最大连续子序列。最大连续前缀序列,最大连续后缀序列,那么每次pushup的时候,依据这3个序列去拼凑得到新的一个结点就可以
代码:
#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;
}