SPOJ GSS5

GSS5 - Can you answer these queries V

#tree

You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| <= 10000 , 1 <= N <= 10000 ). A query is defined as follows: Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 and x1 <= x2 , y1 <= y2 }. Given M queries (1 <= M <= 10000), your program must output the results of these queries.

Input

The first line of the input consist of the number of tests cases <= 5. Each case consist of the integer N and the sequence A. Then the integer M. M lines follow, contains 4 numbers x1, y1, x2 y2.

Output

Your program should output the results of the M queries for each test case, one query per line.

Example

Input:
2
6 3 -2 1 -4 5 2
2
1 1 2 3
1 3 2 5
1 1
1
1 1 1 1

Output:
2
3
1

【题解】
维护前缀/后缀最大，区间最大，区间总和

区间[x1, y1]   [x2, y2]分情况查询
两个区间没有交(y1 < x2), 则答案为[x1, y1].right + [x2, y2].left + (y1 +1 <= x2 - 1 ? [y1 + 1, x2 - 1].sum : 0)
两个区间有交(y1 >= x2), 则分种情况讨论：
              左端点                        右端点
1、左区间非公共部分              公共部分
2、      公共部分　　　　　　 公共部分
3、左区间非公共部分        右区间非公共部分
4、　　公共部分               右区间非公共部分
其实这四种情况可以并为三种：
公共部分最大连续区间
左区间右边最大连续后缀，  右区间非公共部分最大连续前缀
左区间非公共部分最大连续后缀， 右区间最大连续前缀
考虑一下边界，看好怎么+1 -1 合适即可

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #define max(a, b) ((a) > (b) ? (a) : (b))
 #define min(a, b) ((a) < (b) ? (a) : (b))
 //改longlong
 inline void read(long long &x)
 {
     x = ;char ch = getchar(), c = ch;
     while(ch < '' || ch > '')c = ch, ch = getchar();
     while(ch <= '' && ch >= '')x = x *  + ch - '', ch = getchar();
     if(c == '-')x = -x;
 }

 inline void swap(long long &a, long long &b)
 {
     long long tmp = a;a = b;b = tmp;
 }

 const long long MAXN =  + ;
 const long long INF = 0x3f3f3f3f3f3f3f3f;

 long long n,m,num[MAXN],left[MAXN],right[MAXN],ma[MAXN],sum[MAXN];

 void build(long long o = , long long l = , long long r = n)
 {
     if(l == r)
     {
         left[o] = right[o] = ma[o] = sum[o] = num[l];
         return;
     }
     long long mid = (l + r) >> ;
     build(o << , l, mid);
     build(o <<  | , mid + , r);

     left[o] = max(left[o << ], sum[o << ] + left[o <<  | ]);
     right[o] = max(right[o <<  | ], sum[o <<  | ] + right[o <<]);
     ma[o] = max(left[o], max(right[o], left[o <<  | ] + right[o << ]));
     ma[o] = max(ma[o], max(ma[o << ], ma[o <<  | ]));
     sum[o] = sum[o << ] + sum[o <<  | ];
 }

 struct Node
 {
     long long left, right, ma, sum;
     Node(){left = right = ma = -INF;sum = ;}
     Node(long long _left, long long _right, long long _ma, long long _sum){left = _left, right = _right, ma = _ma, sum = _sum;}
 };

 Node ask(long long ll, long long rr, long long o = , long long l = , long long r = n)
 {
     if(ll <= l && rr >= r)return Node(left[o], right[o], ma[o], sum[o]);
     int mid = (l + r) >> ;
     int flag1 = , flag2 = ;
     Node re, ans1, ans2;
     if(mid >= ll) ans1 = ask(ll, rr, o << , l, mid), flag1 = ;
     if(mid < rr)  ans2 = ask(ll, rr, o << | , mid + , r), flag2 = ;

     re.sum = ans1.sum + ans2.sum;
     if(flag1)re.left = max(ans1.left, ans1.sum + ans2.left);
     else re.left = ans2.left;
     if(flag2)re.right = max(ans2.right, ans2.sum + ans1.right);
     else re.right = ans1.right;
     re.ma = max(re.left, max(re.right, max(ans1.right + ans2.left, max(ans1.ma, ans2.ma))));

     return re;
 }

 long long solution(long long x1, long long y1, long long x2, long long y2)
 {
     register Node tmp1, tmp2, tmp3;
     long long ans = -INF;
     if(y1 < x2)
     {
         tmp1 = ask(x1, y1);
         tmp2 = ask(x2, y2);
         if(y1 +  <= x2 - )tmp3 = ask(y1 + , x2 - );
         return tmp1.right + tmp2.left + tmp3.sum;
     }
     else
     {
         tmp1 = ask(x2, y1);
         ans = tmp1.ma;

         tmp2 = ask(x2, y2);
         if(x1 <= x2 - )tmp3 = ask(x1, x2 - );
         else tmp3.right = ;
         ans = max(ans, tmp2.left + tmp3.right);

         tmp2 = ask(x1, y1);
         if(y1 +  <= y2)tmp3 = ask(y1 + , y2);
         else tmp3.left = ;
         ans = max(ans, tmp2.right +  tmp3.left);

         return ans;
     }
 }

 int main()
 {
     long long t;read(t);
     for(;t;--t)
     {
         read(n);
         for(register long long i = ;i <= n;++ i)read(num[i]);
         memset(ma, -0x3f, sizeof(ma));
         memset(left, -0x3f, sizeof(left));
         memset(right, -0x3f, sizeof(right));
         memset(sum, , sizeof(sum));
         build();
         read(m);
         for(register long long i = ;i <= m;++ i)
         {
             long long x1, x2, y1, y2;
             read(x1), read(y1), read(x2), read(y2);
             printf("%lld\n", solution(x1, y1, x2, y2));
         }
     }
     return ;
 } 

SPOJ GSS5

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