解决一个用线段树 [英] Solving a using Segment Tree
问题描述
您给出N的序列A(N< = 50000)-10000和10000之间的整数。在这个序列,你必须使用M(M< = 50000)操作: 修改的第i个元素中的序列,或对于给定的xy打印最大{艾+艾+ 1 + .. + AJ | X - LT; = I< = J< = Y}
You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.
我使用的段树的这一点,但我没有得到正确的输出,请帮我在哪里,我都犯了错
code:
制作树:
I am using Segment Tree for this but i am not getting the correct output , please Help me where i have committed the mistake
CODE:
Making a Tree:
public static void maketree(int current , int a , int b ,int[] arr){
if(b<a) return;
if(b==a) {dp[current] = arr[a]; return ;}
maketree(2*current, a, (a+b)/2, arr);
maketree(2*current+1,1+ (a+b)/2, b, arr);
if(dp[2*current]>0 && dp[2*current+1]>0) dp[current] = dp[2*current] + dp[2*current+1];
else if(dp[2*current]>dp[2*current+1]) dp[current] = dp[2*current];
else dp[current] = dp[2*current+1];
}
更新功能
public static void update(int current , int a , int b , int c , int value){
if(a>b || c<a || c>b) return ;
if(a==b){ dp[current] = value; return ; }
update(2*current, a, (a+b)/2, c, value);
update(2*current+1, (b+a)/2 +1, b, c, value);
if(dp[2*current]>0 && dp[2*current+1]>0) dp[current] = dp[2*current] + dp[2*current+1];
else if(dp[2*current]>dp[2*current+1]) dp[current] = dp[2*current];
else dp[current] = dp[2*current+1];
}
查询功能:
public static int query(int current , int a , int b , int i , int j){
int ans =0;
if(a>j || b<i || a>b) return Integer.MIN_VALUE;
if(a>=i && b<=j) return dp[current];
int x = query(2*current, a, (a+b)/2, i, j);
int y = query(2*current+1, (a+b)/2 +1, b, i, j);
if(x>0 && y>0) ans= x+y;
else if(x>y) ans = x;
else ans =y;
return ans;
}
我不;吨知道我做了错误,请帮忙,将需要什么样的存储容量为 DP阵列
即DP的尺寸
推荐答案
当你合并两个节点,那么它可能会像给below.execute任何简单的例子,让你能感觉到:)
when you are merging two nodes,then it may be like given below.execute any simple example so that you can feel it :)
无效合并(节点,节点B)
{
void merge(node a , node b)
{
sum = a.sum + b.sum;
pre = max(a.pre , (a.sum + b.pre));
suf = max(b.suf , (b.sum + a.suf));
result = max(a.suf + b.pre,max(a.result , b.result));
}
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