• 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
      1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #define max(a, b) ((a) > (b) ? (a) : (b))
  6 #define min(a, b) ((a) < (b) ? (a) : (b)) 
  7 //改longlong 
  8 inline void read(long long &x)
  9 {
 10     x = 0;char ch = getchar(), c = ch;
 11     while(ch < '0' || ch > '9')c = ch, ch = getchar();
 12     while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
 13     if(c == '-')x = -x;
 14 }
 15 
 16 inline void swap(long long &a, long long &b)
 17 {
 18     long long tmp = a;a = b;b = tmp;
 19 }
 20 
 21 const long long MAXN = 100000 + 10;
 22 const long long INF = 0x3f3f3f3f3f3f3f3f;
 23 
 24 long long n,m,num[MAXN],left[MAXN],right[MAXN],ma[MAXN],sum[MAXN];
 25 
 26 void build(long long o = 1, long long l = 1, long long r = n)
 27 {
 28     if(l == r)
 29     {
 30         left[o] = right[o] = ma[o] = sum[o] = num[l];
 31         return;
 32     }
 33     long long mid = (l + r) >> 1;
 34     build(o << 1, l, mid);
 35     build(o << 1 | 1, mid + 1, r);
 36     
 37     left[o] = max(left[o << 1], sum[o << 1] + left[o << 1 | 1]);
 38     right[o] = max(right[o << 1 | 1], sum[o << 1 | 1] + right[o <<1]);
 39     ma[o] = max(left[o], max(right[o], left[o << 1 | 1] + right[o << 1]));
 40     ma[o] = max(ma[o], max(ma[o << 1], ma[o << 1 | 1]));
 41     sum[o] = sum[o << 1] + sum[o << 1 | 1];
 42 }
 43 
 44 struct Node
 45 {
 46     long long left, right, ma, sum;
 47     Node(){left = right = ma = -INF;sum = 0;}
 48     Node(long long _left, long long _right, long long _ma, long long _sum){left = _left, right = _right, ma = _ma, sum = _sum;}
 49 };
 50 
 51 Node ask(long long ll, long long rr, long long o = 1, long long l = 1, long long r = n)
 52 {
 53     if(ll <= l && rr >= r)return Node(left[o], right[o], ma[o], sum[o]);
 54     int mid = (l + r) >> 1;
 55     int flag1 = 0, flag2 = 0;
 56     Node re, ans1, ans2;
 57     if(mid >= ll) ans1 = ask(ll, rr, o << 1, l, mid), flag1 = 1;
 58     if(mid < rr)  ans2 = ask(ll, rr, o <<1 | 1, mid + 1, r), flag2 = 1;
 59     
 60     re.sum = ans1.sum + ans2.sum;
 61     if(flag1)re.left = max(ans1.left, ans1.sum + ans2.left);
 62     else re.left = ans2.left;
 63     if(flag2)re.right = max(ans2.right, ans2.sum + ans1.right);
 64     else re.right = ans1.right;
 65     re.ma = max(re.left, max(re.right, max(ans1.right + ans2.left, max(ans1.ma, ans2.ma))));
 66     
 67     return re;
 68 }
 69 
 70 long long solution(long long x1, long long y1, long long x2, long long y2)
 71 {
 72     register Node tmp1, tmp2, tmp3;
 73     long long ans = -INF;
 74     if(y1 < x2)
 75     {
 76         tmp1 = ask(x1, y1);
 77         tmp2 = ask(x2, y2);
 78         if(y1 + 1 <= x2 - 1)tmp3 = ask(y1 + 1, x2 - 1);
 79         return tmp1.right + tmp2.left + tmp3.sum;
 80     }
 81     else
 82     {
 83         tmp1 = ask(x2, y1);
 84         ans = tmp1.ma;
 85 
 86         tmp2 = ask(x2, y2); 
 87         if(x1 <= x2 - 1)tmp3 = ask(x1, x2 - 1);
 88         else tmp3.right = 0;
 89         ans = max(ans, tmp2.left + tmp3.right);
 90         
 91         tmp2 = ask(x1, y1);
 92         if(y1 + 1 <= y2)tmp3 = ask(y1 + 1, y2);
 93         else tmp3.left = 0;
 94         ans = max(ans, tmp2.right +  tmp3.left);
 95         
 96         return ans;
 97     }
 98 }
 99 
100 int main()
101 {
102     long long t;read(t);
103     for(;t;--t)
104     {
105         read(n);
106         for(register long long i = 1;i <= n;++ i)read(num[i]);
107         memset(ma, -0x3f, sizeof(ma));
108         memset(left, -0x3f, sizeof(left));
109         memset(right, -0x3f, sizeof(right));
110         memset(sum, 0, sizeof(sum));
111         build();
112         read(m);
113         for(register long long i = 1;i <= m;++ i)
114         {
115             long long x1, x2, y1, y2;
116             read(x1), read(y1), read(x2), read(y2);
117             printf("%lld
", solution(x1, y1, x2, y2));
118         } 
119     }
120     return 0;
121 }
    SPOJ GSS5

    注册不了SPJ， 跟标称大数据/小数据（测边界情况）对拍，拍了近半个小时，无错

    

    

    
    
          
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            原文地址：https://www.cnblogs.com/huibixiaoxing/p/7487975.html
          
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