操作：
1，更新某个位置的值。
2，查询区间最大连续和。
更新位置然后PushUp。
查询同【SPOJ】1043 Can you answer these queries I。
1 #include<cstdio> 2 #define MAX(a,b) ((a)>(b)?(a):(b)) 3 #define MAXN 50010 4 #define oo 1000000000 5 struct node { 6 int left, right, sum, val; 7 void Init() { 8 sum = 0; 9 left = right = val = -oo; 10 } 11 }; 12 node tree[MAXN << 2]; 13 inline void PushUp(int rt) { 14 tree[rt].left = MAX(tree[rt << 1].left, 15 tree[rt << 1].sum + tree[rt << 1 | 1].left); 16 tree[rt].right = MAX(tree[rt << 1 | 1].right, 17 tree[rt << 1 | 1].sum + tree[rt << 1].right); 18 tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum; 19 tree[rt].val = MAX(tree[rt << 1].val, tree[rt << 1 | 1].val); 20 tree[rt].val = MAX(tree[rt].val, 21 tree[rt << 1].right + tree[rt << 1 | 1].left); 22 } 23 void Build(int L, int R, int rt) { 24 if (L == R) { 25 scanf("%d", &tree[rt].val); 26 tree[rt].left = tree[rt].right = tree[rt].sum = tree[rt].val; 27 } else { 28 int mid = (L + R) >> 1; 29 Build(L, mid, rt << 1); 30 Build(mid + 1, R, rt << 1 | 1); 31 PushUp(rt); 32 } 33 } 34 void Update(int x, int val, int L, int R, int rt) { 35 if (L == R) 36 tree[rt].left = tree[rt].right = tree[rt].sum = tree[rt].val = val; 37 else { 38 int mid = (L + R) >> 1; 39 if (x <= mid) 40 Update(x, val, L, mid, rt << 1); 41 else 42 Update(x, val, mid + 1, R, rt << 1 | 1); 43 PushUp(rt); 44 } 45 } 46 node Query(int x, int y, int L, int R, int rt) { 47 if (x <= L && R <= y) 48 return tree[rt]; 49 int mid = (L + R) >> 1; 50 node a, b, res; 51 a.Init(), b.Init(), res.Init(); 52 if (x <= mid) 53 a = Query(x, y, L, mid, rt << 1); 54 if (y > mid) 55 b = Query(x, y, mid + 1, R, rt << 1 | 1); 56 res.left = MAX(a.left, a.sum + b.left); 57 res.right = MAX(b.right, b.sum + a.right); 58 res.sum = a.sum + b.sum; 59 res.val = MAX(a.val, b.val); 60 res.val = MAX(res.val, a.right + b.left); 61 return res; 62 } 63 int main() { 64 int n, q, cmd, x, y; 65 while (~scanf("%d", &n)) { 66 Build(1, n, 1); 67 scanf("%d", &q); 68 while (q--) { 69 scanf("%d%d%d", &cmd, &x, &y); 70 if (cmd) 71 printf("%d\n", Query(x, y, 1, n, 1).val); 72 else 73 Update(x, y, 1, n, 1); 74 } 75 } 76 return 0; 77 }