【SPOJ】2916 Can you answer these queries V

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 }