Exemplo n.º 1
0
 int maxSum(TreeNode *root, int *maxChild) {
     int lm = 0;
     int rm = 0;
     
     *maxChild = 0;
     if (!root) return 0;
     int left = maxSum(root->left, &lm);
     int right = maxSum(root->right, &rm);
     
     int a = max(lm, rm);
     *maxChild = root->val;
     if (a > 0)
         *maxChild += a;
     
     a = 0;
     if (lm > 0)
         a += lm;
     if (rm > 0)
         a += rm;
     a += root->val;
     
     if (root->left)
         a = max(a, left);
     if (root->right)
         a = max(a, right);
     
     return a;
 }
Exemplo n.º 2
0
int maxSum(struct node* root, int* max)
{
    if(root == NULL)
        return 0;
    int sum = maxSum(root->left, max) + maxSum(root->right, max) + root->data;
    if(sum > *max)
        *max = sum;
    return sum;
}
 int maxSum(TreeNode* root) {
   if (root == NULL)
     return 0;
   int leftSum = maxSum(root->left);
   int rightSum = maxSum(root->right);
   if (leftSum + rightSum + root->val > r) {
     r = leftSum + rightSum + root->val;
   }
   return max(0, max(leftSum, rightSum) + root->val);
 }
Exemplo n.º 4
0
void pull( Treap * a ){
	a->sum = Sum( a->l ) + Sum( a->r ) + a->val;
	a->lsum = Sum( a->l ) + a->val + max( 0 , lSum( a->r ) );
	if( a->l ) a->lsum = max( lSum( a->l ) , a->lsum );
	a->rsum = Sum( a->r ) + a->val + max( 0 , rSum( a->l ) );
	if( a->r ) a->rsum = max( rSum( a->r ) , a->rsum );
	a->maxsum = max( 0 , rSum( a->l ) ) + a->val + max( 0 , lSum( a->r ) );
	a->maxsum = max( a->maxsum , max( maxSum( a->l ) , maxSum( a->r ) ) );
	a->sz = Size( a->l ) + Size( a->r ) + 1;
}
Exemplo n.º 5
0
int maxSum(Node *root){
	if(!root)
		return 0;
	int leftSum = maxSum(root->left);
	int rightSum = maxSum(root->right);
	
	int max =( (leftSum > rightSum) ? leftSum:rightSum );
	
	return (max + root->data);
}
//函数返回当前节点为根的最大和
int maxSum(TreeNode* root,int &sum){
    if(root==NULL)
        return INT_MIN;
    sum=root->val>sum?root->val:sum;
    int leftSum=maxSum(root->left,sum);
    int rightSum=maxSum(root->right,sum);
    leftSum=leftSum>0?leftSum:0;  //若左子树和为负数,就直接设为0
    rightSum=rightSum>0?rightSum:0; //和左子树相同
    sum=max(leftSum+root->val+rightSum,sum);
    return root->val+max(leftSum,rightSum);
}
 int maxSum(TreeNode* root) {
     if(!root) return 0;
     if(!root->left && !root->right) {
         max_global = std::max(max_global,root->val);
         return root->val;
     }
     int left_val = maxSum(root->left);
     int right_val = maxSum(root->right); 
     int max_local = std::max(root->val, std::max(left_val+root->val, right_val+root->val));
     max_global = std::max(max_local, std::max(left_val+right_val+root->val, max_global));
     return max_local;
 }
Exemplo n.º 8
0
 int maxSum(TreeNode* root,int& mymax){  
     if(!root)  
         return 0;  
     int left=maxSum(root->left,mymax);  
     int right=maxSum(root->right,mymax);  
     int localmax=root->val;  
     if(left>0)  
         localmax+=left;  
     if(right>0)  
         localmax+=right;  
     mymax=max(mymax,localmax);  
     return max(root->val,max(root->val+left,root->val+right));  
 }  
int maxSum(TreeNode *root)
{
	if (!root) {
		return 0;
	}

	int l = maxSum(root->left), r = maxSum(root->right);

	int cur_max = (l > 0 ? l : 0) + (r > 0 ? r : 0) + root->val;
	if (cur_max > ans) {
		ans = cur_max;
	}
	
	return max(max(l, r) + root->val, 0);
}
 // Value may be negative!!
 int maxPathSum(TreeNode* root) {
   // Start typing your C/C++ solution below
   // DO NOT write int main() function
   r = INT_MIN;
   maxSum(root);
   return r;
 }
Exemplo n.º 11
0
int main()
{
	int array[10] = {1, -2, 3, 4, -5, 6, -7, 8, 9, 10}; 
	int max = maxSum(array, sizeof(array)/sizeof(int));
	printf("max sum of array is : %d\n", max);
	return 0;
}
Exemplo n.º 12
0
 int maxPathSum(TreeNode *root) {  
     // Start typing your C/C++ solution below  
     // DO NOT write int main() function  
     int mymax=root->val;  
     int val=maxSum(root,mymax);  
     return max(mymax,val);  
 }  
int main()  
{  
    int a[10]={1, -2, 3, 10, -4, 7, 2, -5};  
    //int a[]={-1,-2,-3,-4};  //测试全是负数的用例  
    printf("子数组的最大和为:%d\n",maxSum(a,8));
    return 0;  
}  
Exemplo n.º 14
0
int main()
{ 
  int i,N,T;
  int tmp = 0;
  scanf("%d",&T);
  while(T > 0)
  {
    scanf("%d",&N);
    for(i = 0;i < N;i++)
    {
      scanf("%d",&a[i]);
    }
    T--;
//    for(i = 0;i < N;i++)
//    {
//      printf("%d ",a[i]);
//    }
//    printf("\n");
    printf("Case %d:\n",++tmp);
    printf("%d",maxSum(a,N));
    printf(" %d %d\n",first,last);
    if(T > 0)
      printf("\n");
  }
  return 0;
}
Exemplo n.º 15
0
// kadane's algo!
int main()
{
	int a[] = {-2,1,-3,4,-1,2,1,-5,4};
	int i = maxSum(a, 9);
	printf("%d\n", i);
	return 0;
}
Exemplo n.º 16
0
 int maxPathSum(TreeNode *root) {
     // Start typing your C/C++ solution below
     // DO NOT write int main() function
     //if (!root) return 0;
     int c = 0;
     int r = maxSum(root, &c);
     return r;
 }
Exemplo n.º 17
0
int main()
{
    int a[]={1,5,9,8,2,2,4,1},maxsum;
    maxsum=maxSum(a,sizeof(a)/sizeof(a[0]));
    printf("max sum is: %d",maxsum);
    getch();
        
    
}
Exemplo n.º 18
0
void solve(){
	Treap *t = NULL , *tl = NULL , *tr = NULL;
	n = getint(); m = getint();
	for( int i = 0 ; i < n ; i ++ )
		t = merge( t , new Treap( getint() ) );
	while( m -- ){
		scanf( "%s" , c );
		if( c[ 0 ] == 'I' ){
			int p , k;
			p = getint(); k = getint();
			split( t , p , tl , tr );
			t = NULL;
			while( k -- )
				t = merge( t , new Treap( getint() ) );
			t = merge( t , tr );
			t = merge( tl , t );
		}else if( c[ 0 ] == 'D' ){
			int p , k;
			p = getint(); k = getint();
			split( t , p - 1 , tl , t );
			split( t , k , t , tr );
			Delete( t );
			t = merge( tl , tr );
		}else if( c[ 0 ] == 'R' ){
			int p , k;
			p = getint(); k = getint();
			split( t , p - 1 , tl , t );
			split( t , k , t , tr );
			t->tag ^= 1;
			int swp = t->lsum; t->lsum = t->rsum; t->rsum = swp;
			t = merge( t , tr );
			t = merge( tl , t );
		}else if( c[ 0 ] == 'G' ){
			int p , k;
			p = getint(); k = getint();
			split( t , p - 1 , tl , t );
			split( t , k , t , tr );
			printf( "%d\n" , Sum( t ) );
			t = merge( t , tr );
			t = merge( tl , t );
		}else if( c[ 2 ] == 'K' ){
			int p , k;
			p = getint(); k = getint();
			split( t , p - 1 , tl , t );
			split( t , k , t , tr );
			t->tagn = true; t->num = getint();
			t->sum = t->num * t->sz;
			if( t->num >= 0 )
				t->lsum = t->rsum = t->maxsum = t->sum;
			else t->lsum = t->rsum = t->maxsum = t->num;
			t = merge( t , tr );
			t = merge( tl , t );
		}else printf( "%d\n" , maxSum( t ) );

	}
}
int maxPathSum(TreeNode *root)
{
	if (!root) {
		return 0;
	}

	maxSum(root);

	return ans;
}
	int maxSum(TreeNode * head)
	{
		//返回的时候只能返回一条边给上层
		//但是计算当前node的时候,需要把两条边算进去。
	    int ret = INT_MIN;
		int left = 0, right = 0;
		if (head->left)
			left = maxSum(head->left);
		if (head->right)
			right = maxSum(head->right);
		int curMax = (left > 0 ? left : 0) + (right > 0 ? right : 0) + head->val;
		int m = left > right?left:right;
		if(m > 0 )
			ret = m + head->val;
		else
			ret = head->val;
		sum = max(curMax, sum);
		return  ret;
	}
int maxSum(int left, int right)
{
  int mid, sum, maxToLeft, maxToRight, maxCrossing, maxInA, maxInB;
  int i;

  // Zero element Vector
  if (left>right) {
      return 0;
  }

  // One element Vector
  if (left==right) {
      return max(0,list[left]);
  }

  // A is List[Left...Mid], B is List[Mid+1...Right]
  mid = (left + right)/2;

  // Find max crossing to left
  sum=0;
  maxToLeft=0;
  for(i=mid; i>=left; i--)
  {
     sum = sum + list[i];
     maxToLeft = max(maxToLeft,sum);
  }

  // Find max crossing to right
  sum=0;
  maxToRight=0;
  for(i=mid+1; i<=right; i++)
  {
      sum = sum + list[i];
      maxToRight = max(maxToRight,sum);
  }

  maxCrossing = maxToLeft + maxToRight;
  maxInA = maxSum(left,mid);
  maxInB = maxSum(mid+1,right);

 return max(max(maxCrossing,maxInA),maxInB);
}
Exemplo n.º 22
0
int main(){
    struct node *root = NULL;
 
    /* Constructing tree given in the above figure */
    root = newNode(10);
    root->left = newNode(-2);
    root->right = newNode(7);
    root->left->left = newNode(8);
    root->left->right = newNode(-4);
	
    std::cout<<"max root to leaf sum:"<<maxSum(root)<<std::endl;
    return 0;
}
Exemplo n.º 23
0
int main() {
    scanf("%d",&t);
    for(i=1; i<=t; i++) {
        init();
        scanf("%d",&n);
        for(j=0; j<n; j++) {
            scanf("%d",&a[j]);
        }
        maxSum(a);
        printf("Case %d:\n",i);
        printf("%d %d %d\n",max,lflag,rflag);
        if(i!=t)printf("\n");
    }
    return 0;
}
Exemplo n.º 24
0
int main()
{
  struct node *root = newNode(1);
  root->left        = newNode(-10);
  root->right       = newNode(-30);
  root->left->left  = newNode(4);
  root->left->right = newNode(5);
  root->left->left->left  = newNode(-6);


    int max=0;

    maxSum(root, &max);
    printf("maxSum = %d\n", max);

    return 0;
}
Exemplo n.º 25
0
int main()
{
	int a[SIZE];
	int n = sizeof(a) / sizeof(a[0]);	// 배열 원소의 갯수
	clock_t t1, t2, t3, t4;

	init(a, n);
	t1 = clock();
	printf("maxSum			: %d\n", maxSum(a, n));
	t2 = clock();
	printf("fastMaxSum		: %d\n", fastMaxSum(a, n));
	t3= clock();
	printf("fastestMaxSum	: %d\n", fastestMaxSum(a, n));	
	t4 = clock();
	// 수행시간 측정
	printf("maxSum			: %lf\n", (double)(t2 - t1) / CLOCKS_PER_SEC);
	printf("fastMaxSum		: %lf\n", (double)(t3 - t2) / CLOCKS_PER_SEC);
	printf("fastestMaxSum	: %lf\n", (double)(t4 - t3) / CLOCKS_PER_SEC);
	return 0;
}
int main(int argc, char** argv) {

    printf ("%d \n", maxSum(0,4));

    return 0;
}
int maxPathSum(TreeNode* root) {
    int sum=INT_MIN; //保存最大值
    maxSum(root,sum);
    return sum;
}
	int maxPathSum(TreeNode *root)
	{
		sum = root->val;
		maxSum(root);
		return sum;
	}
Exemplo n.º 29
0
/////implement APcluser algorithm
void APcluster(float s[][R],float p[],int idx[])
{ 
    float Max(float a[][R],float s[][R],int i);
    float maxSum(float z,float r[][R],int k);
    float lam=L;//¦Ë
    int i,j,k,k1;
    int times=0,mx=0;
	float a[R][R],r[R][R],e[R][R];
	float sum=0.0,tem=0.0,old=0.0,Rdis=0.0,Rdsum=0.0,Rdaver=0.0,Adis=0.0,Adsum=0.0,Adaver=0.0;
	bool flag=true;
     //////Initialize///
	for(i=0;i<R;i++)
	{
		for(j=0;j<R;j++)
		{
		   a[i][j]=0.0;
		   r[i][j]=0.0;
		   e[i][j]=0.0;
        }	   
	}
	/////iterations////
	while(flag)
	{
	    times++;
	    Rdsum=0.0;
	    Adsum=0.0;
	    ///Responsibility
       for(i=0;i<R;i++) 
       {
          k=0;
          for(k=0;k<R;k++) 
          {
              old=r[i][k];
              r[i][k]=s[i][k]-Max(a,s,i);       //formula1:r(i,k)<-s(i,k)-max{a(i,k')+s(i,k')}
              r[i][k]=(1-lam)*old+lam*r[i][k];  //formula2:r(new)=¦Ë*r(new)+(1-¦Ë)*r(old)
              Rdis=r[i][k]-old; //calculate the change 
             // printf("Rdis(%d,%d):%f ",i,k,Rdis);
              Rdsum=Rdsum+Rdis; //sum of the changes
          }             
       }   
       Rdaver=Rdsum/R;  //calculate the average of the sum of the changes 
          //printf("\nRdaver %f , ",Rdaver); //test
         ///availabilities 
        for(j=0;j<R;j++) 
       {
          k1=0;
          while(k1<R)
          {
              old=a[j][k1];
              if(j==k1)
              {
                  a[j][k1]=maxSum(0.0,r,k1);
              }    
              else
              {
                  tem=r[k1][k1]+maxSum(0.0,r,k1);  //formula3: r(k,k)+sum of max{0,r(i',k)}
                  a[j][k1]=(0.0<tem)?0.0:tem;      //formula4:min{0,formula3}
              }    
              
              a[j][k1]=(1-lam)*old+lam*a[j][k1];  //formula:r(new)=¦Ë*r(new)+(1-¦Ë)*r(old)
              Adis=a[j][k1]-old;  //calculate the change 
              Adsum=Adsum+Adis;   //sum of the changes
              k1++;
          }   //while                   
       }  //for
       
        Adaver=Adsum/R;  //calculate the average of the sum of the changes  
		
	  for(i=0;i<R;i++)
     {
   
         for(j=0;j<R;j++)
         {
             e[i][j]=a[i][j]+r[i][j];

         }  
         
     }   
     //e(i,k)=maxk{e(i,k)},then make the exemplar decisions
	  for(i=0;i<R;i++)
     {
         mx=0;  
        
        for(j=0;j<R;j++)
         {
             if(e[i][j]>e[i][mx])
             {
               	  mx=j;
             }  
		 }  
		 	idx[i]=mx;  
     }     
        //printf("\nAdaver %f , ",Adaver); //test
        if(((Rdaver*Rdaver)<Threshold) && ((Adaver*Adaver) <Threshold)||(times>MAXI))	
          flag=false;
               
     }
     
printf("Iterators : %d\n ",times);
} 
 int maxPathSum(TreeNode* root) {
     max_global = root->val;
     maxSum(root);
     return max_global;
 }