SPOJ GSS1 Can you answer these queries I
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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。
#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;}
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