操作：
给出x属于[x1,y1]，y属于[x2,y2]，求[x,y]的最大连续和。
将区间分3段考虑，答案可能由某一段的最大连续和，或者某一段往左的最大连续和，某一段往右的最大连续和组合而来。需要特判的是，区间可能存在包含关系。
1 #include<cstdio> 2 #define MAX(a,b) ((a)>(b)?(a):(b)) 3 #define MAXN 10010 4 #define oo 987654321 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].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum; 15 tree[rt].left = MAX(tree[rt<<1].left,tree[rt<<1].sum+tree[rt<<1|1].left); 16 tree[rt].right = 17 MAX(tree[rt<<1|1].right,tree[rt<<1|1].sum+tree[rt<<1].right); 18 tree[rt].val = MAX(tree[rt<<1].val,tree[rt<<1|1].val); 19 tree[rt].val = MAX(tree[rt].val,tree[rt<<1].right+tree[rt<<1|1].left); 20 } 21 void Build(int L, int R, int rt) { 22 if (L == R) { 23 scanf("%d", &tree[rt].val); 24 tree[rt].left = tree[rt].right = tree[rt].sum = tree[rt].val; 25 } else { 26 int mid = (L + R) >> 1; 27 Build(L, mid, rt << 1); 28 Build(mid + 1, R, rt << 1 | 1); 29 PushUp(rt); 30 } 31 } 32 node Query(int x, int y, int L, int R, int rt) { 33 if (x <= L && R <= y) 34 return tree[rt]; 35 int mid = (L + R) >> 1; 36 node a, b, res; 37 a.Init(), b.Init(); 38 if (x <= mid) 39 a = Query(x, y, L, mid, rt << 1); 40 if (y > mid) 41 b = Query(x, y, mid + 1, R, rt << 1 | 1); 42 res.left = MAX(a.left,a.sum+b.left); 43 res.right = MAX(b.right,b.sum+a.right); 44 res.sum = a.sum + b.sum; 45 res.val = MAX(a.val,b.val); 46 res.val = MAX(res.val,a.right+b.left); 47 return res; 48 } 49 int main() { 50 int c, n, q, ans, tmp1, tmp2, x1, y1, x2, y2; 51 scanf("%d", &c); 52 while (c--) { 53 scanf("%d", &n); 54 Build(1, n, 1); 55 scanf("%d", &q); 56 while (q--) { 57 scanf("%d%d%d%d", &x1, &y1, &x2, &y2); 58 if (y1 <= x2) { 59 ans = 0; 60 tmp1 = tmp2 = -oo; 61 if (y1 + 1 <= x2 - 1) 62 ans += Query(y1 + 1, x2 - 1, 1, n, 1).sum; 63 if (y1 == x2) { 64 tmp1 = Query(x2, y2, 1, n, 1).left; 65 ans += tmp1; 66 if (x1 <= y1 - 1) { 67 tmp2 = Query(x1, y1 - 1, 1, n, 1).right; 68 ans += tmp2; 69 } 70 ans = MAX(ans,tmp1); 71 tmp2 = Query(x1, y1, 1, n, 1).right; 72 ans = MAX(ans,tmp2); 73 } else 74 ans += Query(x2, y2, 1, n, 1).left 75 + Query(x1, y1, 1, n, 1).right; 76 } else { 77 ans = Query(x2, y1, 1, n, 1).val; 78 tmp1 = Query(y1, y2, 1, n, 1).left 79 + Query(x1, x2, 1, n, 1).right; 80 if (x2 + 1 <= y1 - 1) 81 tmp1 += Query(x2 + 1, y1 - 1, 1, n, 1).sum; 82 ans = MAX(ans,tmp1); 83 tmp1 = Query(x1, x2, 1, n, 1).right; 84 if (y1 >= x2) 85 ans = MAX(ans,tmp1); 86 if (x2 < y2) 87 tmp1 += Query(x2 + 1, y2, 1, n, 1).left; 88 ans = MAX(ans,tmp1); 89 tmp1 = Query(y1, y2, 1, n, 1).left; 90 if (x2 <= y1) 91 ans = MAX(ans,tmp1); 92 if (x1 < y1) 93 tmp1 += Query(x1, y1 - 1, 1, n, 1).right; 94 ans = MAX(ans,tmp1); 95 } 96 printf("%d\n", ans); 97 } 98 } 99 return 0; 100 }