Can you answer these queries V 【SPOJ - GSS5】【线段树 最大子段和】

题目链接

  WA了好多发，最后的时候把两个其中两个分开的判断「y1 == y2 与 else」放到一块去之后尽然直接A了。

  有一个整数序列a
。有m次询问，每次询问Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 , x1 <= x2 , y1 <= y2 }。

  然后，就是去考虑怎样处理这个关系，直接分类讨论吧讨论区间是分开的y1 < y2这样的类型，还是else。

给两组样例：

[code]1
6
12 -6 -5 -8 -5 -4
1
1 2 2 3

[code]1
6
12 -6 -5 -8 -5 -4
1
1 3 2 5

[code]#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define efs 1e-6
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define max3(a, b, c) max(a, max(b, c))
#define max4(a, b, c, d) max(max(a, b), max(c, d))
#define max5(a, b, c, d, f) max4(a, b, c, max(d, f))
#define max6(a, b, c, d, f, g) max(max3(a, b, c), max3(d, f, g))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e4 + 7;
int N, M;
struct node
{
int lm, rm, mx, sum;
node(int a=0, int b=0, int c=0, int d=0):lm(a), rm(b), mx(c), sum(d) {}
void give(int x) { lm = rm = mx = sum = x; }
}tree[maxN<<2];
inline void pushup(node &root, node lc, node rc)
{
root.lm = max(lc.lm, lc.sum + rc.lm);
root.rm = max(rc.rm, rc.sum + lc.rm);
root.sum = lc.sum + rc.sum;
root.mx = max5(root.lm, root.rm, lc.mx, rc.mx, lc.rm + rc.lm);
}
void buildTree(int rt, int l, int r)
{
if(l == r)
{
int val;
scanf("%d", &val);
tree[rt].give(val);
return;
}
int mid = HalF;
buildTree(Lson); buildTree(Rson);
pushup(tree[rt], tree[lsn], tree[rsn]);
}
node query(int rt, int l, int r, int ql, int qr)
{
if(ql <= l && qr >= r) return tree[rt];
int mid = HalF;
if(qr <= mid) return query(QL);
else if(ql > mid) return query(QR);
else
{
node TL = query(QL), TR = query(QR);
node ans;   pushup(ans, TL, TR);
return ans;
}
}
int main()
{
int Cas; scanf("%d", &Cas);
while(Cas--)
{
scanf("%d", &N);
buildTree(1, 1, N);
scanf("%d", &M);
int range_left = 0, range_right = 0, range_sum = 0;
for(int i=1, x1, y1, x2, y2; i<=M; i++)
{
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
if(y1 < x2)     //x1 <= y1 < x2 <= y2
{
range_left = query(1, 1, N, x1, y1).rm;
range_right = query(1, 1, N, x2, y2).lm;
range_sum = 0;
if(x2 - y1 > 1) range_sum = query(1, 1, N, y1 + 1, x2 - 1).sum;
printf("%d\n", range_left + range_right + range_sum);
}
else
{
int ans;
range_left = query(1, 1, N, x1, x2).rm;
range_right = 0;
if(x2 < y2) range_right = query(1, 1, N, x2 + 1, y2).lm;
range_right = max(0, range_right);
ans = range_left + range_right;
range_left = query(1, 1, N, x1, y1).rm;
range_right = 0;
if(y1 < y2) range_right = query(1, 1, N, y1 + 1, y2).lm;
ans = max(ans, range_left + range_right);
ans = max(ans, query(1, 1, N, x2, y1).mx);
printf("%d\n", ans);
}
}
}
return 0;
}