SPOJ GSS1 Can you answer these queries I
2015-04-09 18:50
417 查看
题意: 给定一个序列和m个询问, 要求求出一个区间内的最大连续子段和。
思路:
用线段树维护一个区间左连续最大子段和,右连续最大子段和,区间和,区间最大连续子段和这4个东西, 然后就可以搞了。。
查询的时候,如果查询的区间过当前区间的中点的话,就要查询左边区间的右连续最大值和右边区间左连续的最大值, 加起来和答案比较。
思路:
用线段树维护一个区间左连续最大子段和,右连续最大子段和,区间和,区间最大连续子段和这4个东西, 然后就可以搞了。。
查询的时候,如果查询的区间过当前区间的中点的话,就要查询左边区间的右连续最大值和右边区间左连续的最大值, 加起来和答案比较。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; #define mxn 100020 #define ls (i << 1) #define rs (ls | 1) #define md ((ll + rr) >> 1) int lx[mxn<<2], rx[mxn<<2], mx[mxn<<2], sum[mxn<<2]; int n, a[mxn]; void push_up(int i) { sum[i] = sum[ls] + sum[rs]; mx[i] = max(mx[ls], mx[rs]); mx[i] = max(mx[i], rx[ls] + lx[rs]); lx[i] = max(lx[ls], sum[ls] + lx[rs]); rx[i] = max(rx[rs], sum[rs] + rx[ls]); } void build(int ll, int rr, int i) { if(ll == rr) { scanf("%d", &mx[i]); sum[i] = lx[i] = rx[i] = mx[i]; return; } build(ll, md, ls), build(md + 1, rr, rs); push_up(i); } int ql(int l, int r, int ll, int rr, int i) { if(ll == l && rr == r) return lx[i]; if(r <= md) return ql(l, r, ll, md, ls); if(l > md) return ql(l, r, md + 1, rr, rs); return max(lx[ls], sum[ls] + ql(md + 1, r, md + 1, rr, rs)); } int qr(int l, int r, int ll, int rr, int i) { if(ll == l && rr == r) return rx[i]; if(r <= md) return qr(l, r, ll, md, ls); if(l > md) return qr(l, r, md + 1, rr, rs); return max(rx[rs], sum[rs] + qr(l, md, ll, md, ls)); } int query(int l, int r, int ll, int rr, int i) { if(ll == l && rr == r) return mx[i]; if(r <= md) return query(l, r, ll, md, ls); if(l > md) return query(l, r, md + 1, rr, rs); int LX = ql(md + 1, r, md + 1, rr, rs); int RX = qr(l, md, ll, md, ls); LX += RX; return max(LX, max(query(l, md, ll, md, ls), query(md + 1, r, md + 1, rr, rs))); } int main() { while(scanf("%d", &n) != EOF) { build(1, n, 1); int m; scanf("%d", &m); while(m--) { int l, r; scanf("%d%d", &l, &r); printf("%d\n", query(l, r, 1, n, 1)); } } return 0; }
相关文章推荐
- 【线段树】spoj GSS2 Can you answer these queries II
- SPOJ GSS2 Can you answer these queries II
- SPOJ GSS6 4487. Can you answer these queries VI (SPLAY)
- SPOJ GSS3 Can you answer these queries III ——线段树
- SPOJ GSS1 Can you answer these queries I(区间合并)
- SPOJ GSS6 Can you answer these queries VI
- SPOJ GSS1_Can you answer these queries I(线段树区间合并)
- SPOJ GSS1 Can you answer these queries I[线段树]
- SPOJ GSS3 Can you answer these queries III
- BZOJ 2482 || SPOJ GSS2 Can you answer these queries II(线段树 离线 后缀和)
- SPOJ 2916 Can you answer these queries V(GSS5 线段树)
- spoj 1557 Can you answer these queries II (gss2)线段树
- spojGSS2 1557 Can you answer these queries II(成段更新)
- SPOJ GSS3 Can you answer these queries III[线段树]
- SPOJ GSS3 Can you answer these queries III
- SPOJ GSS2 - Can you answer these queries II(线段树 区间修改+区间查询)(后缀和)
- GSS5 spoj 2916. Can you answer these queries V 线段树
- SPOJ GSS6 Can you answer these queries VI ——Splay
- SPOJ GSS1 Can you answer these queries I
- SPOJ - GSS4 Can you answer these queries IV