pqdownheap() 개인적인 메소드

private pqdownheap ( short tree, int k ) : void
tree short
k int
리턴 void
예제 #1
0
        internal void build_tree(Deflate s)
        {
            short[] array       = dyn_tree;
            short[] static_tree = stat_desc.static_tree;
            int     elems       = stat_desc.elems;
            int     num         = -1;

            s.heap_len = 0;
            s.heap_max = 573;
            for (int i = 0; i < elems; i++)
            {
                if (array[i * 2] != 0)
                {
                    num        = (s.heap[++s.heap_len] = i);
                    s.depth[i] = 0;
                }
                else
                {
                    array[i * 2 + 1] = 0;
                }
            }
            int num2;

            while (s.heap_len < 2)
            {
                num2            = (s.heap[++s.heap_len] = ((num < 2) ? (++num) : 0));
                array[num2 * 2] = 1;
                s.depth[num2]   = 0;
                s.opt_len--;
                if (static_tree != null)
                {
                    s.static_len -= static_tree[num2 * 2 + 1];
                }
            }
            max_code = num;
            for (int i = s.heap_len / 2; i >= 1; i--)
            {
                s.pqdownheap(array, i);
            }
            num2 = elems;
            do
            {
                int i = s.heap[1];
                s.heap[1] = s.heap[s.heap_len--];
                s.pqdownheap(array, 1);
                int num3 = s.heap[1];
                s.heap[--s.heap_max] = i;
                s.heap[--s.heap_max] = num3;
                array[num2 * 2]      = (short)(array[i * 2] + array[num3 * 2]);
                s.depth[num2]        = (byte)(Math.Max(s.depth[i], s.depth[num3]) + 1);
                array[i * 2 + 1]     = (array[num3 * 2 + 1] = (short)num2);
                s.heap[1]            = num2++;
                s.pqdownheap(array, 1);
            }while (s.heap_len >= 2);
            s.heap[--s.heap_max] = s.heap[1];
            gen_bitlen(s);
            gen_codes(array, num, s.bl_count);
        }
        // Construct one Huffman tree and assigns the code bit strings and lengths.
        // Update the total bit length for the current block.
        // IN assertion: the field freq is set for all tree elements.
        // OUT assertions: the fields len and code are set to the optimal bit length
        //     and corresponding code. The length opt_len is updated; static_len is
        //     also updated if stree is not null. The field max_code is set.
        internal void build_tree(Deflate s){
            short[] tree=dyn_tree;
            short[] stree=stat_desc.static_tree;
            int elems=stat_desc.elems;
            int n, m;          // iterate over heap elements
            int max_code=-1;   // largest code with non zero frequency
            int node;          // new node being created

            // Construct the initial heap, with least frequent element in
            // heap[1]. The sons of heap[n] are heap[2*n] and heap[2*n+1].
            // heap[0] is not used.
            s.heap_len = 0;
            s.heap_max = HEAP_SIZE;

            for(n=0; n<elems; n++) {
                if(tree[n*2] != 0) {
                    s.heap[++s.heap_len] = max_code = n;
                    s.depth[n] = 0;
                }
                else{
                    tree[n*2+1] = 0;
                }
            }

            // The pkzip format requires that at least one distance code exists,
            // and that at least one bit should be sent even if there is only one
            // possible code. So to avoid special checks later on we force at least
            // two codes of non zero frequency.
            while (s.heap_len < 2) {
                node = s.heap[++s.heap_len] = (max_code < 2 ? ++max_code : 0);
                tree[node*2] = 1;
                s.depth[node] = 0;
                s.opt_len--; if (stree!=null) s.static_len -= stree[node*2+1];
                // node is 0 or 1 so it does not have extra bits
            }
            this.max_code = max_code;

            // The elements heap[heap_len/2+1 .. heap_len] are leaves of the tree,
            // establish sub-heaps of increasing lengths:

            for(n=s.heap_len/2;n>=1; n--)
                s.pqdownheap(tree, n);

            // Construct the Huffman tree by repeatedly combining the least two
            // frequent nodes.

            node=elems;                 // next internal node of the tree
            do{
                // n = node of least frequency
                n=s.heap[1];
                s.heap[1]=s.heap[s.heap_len--];
                s.pqdownheap(tree, 1);
                m=s.heap[1];                // m = node of next least frequency

                s.heap[--s.heap_max] = n; // keep the nodes sorted by frequency
                s.heap[--s.heap_max] = m;

                // Create a new node father of n and m
                tree[node*2] = (short)(tree[n*2] + tree[m*2]);
                s.depth[node] = (byte)(System.Math.Max(s.depth[n],s.depth[m])+1);
                tree[n*2+1] = tree[m*2+1] = (short)node;

                // and insert the new node in the heap
                s.heap[1] = node++;
                s.pqdownheap(tree, 1);
            }
            while(s.heap_len>=2);

            s.heap[--s.heap_max] = s.heap[1];

            // At this point, the fields freq and dad are set. We can now
            // generate the bit lengths.

            gen_bitlen(s);

            // The field len is now set, we can generate the bit codes
            gen_codes(tree, max_code, s.bl_count);
        }
예제 #3
0
        // Construct one Huffman tree and assigns the code bit strings and lengths.
        // Update the total bit length for the current block.
        // IN assertion: the field freq is set for all tree elements.
        // OUT assertions: the fields len and code are set to the optimal bit length
        //     and corresponding code. The length opt_len is updated; static_len is
        //     also updated if stree is not null. The field max_code is set.
        internal void build_tree(Deflate s)
        {
            short[] tree = dyn_tree;
            short[] stree = stat_desc.static_tree;
            int     elems = stat_desc.elems;
            int     n, m;          // iterate over heap elements
            int     max_code = -1; // largest code with non zero frequency
            int     node;          // new node being created

            // Construct the initial heap, with least frequent element in
            // heap[1]. The sons of heap[n] are heap[2*n] and heap[2*n+1].
            // heap[0] is not used.
            s.heap_len = 0;
            s.heap_max = HEAP_SIZE;

            for (n = 0; n < elems; n++)
            {
                if (tree[n * 2] != 0)
                {
                    s.heap[++s.heap_len] = max_code = n;
                    s.depth[n]           = 0;
                }
                else
                {
                    tree[n * 2 + 1] = 0;
                }
            }

            // The pkzip format requires that at least one distance code exists,
            // and that at least one bit should be sent even if there is only one
            // possible code. So to avoid special checks later on we force at least
            // two codes of non zero frequency.
            while (s.heap_len < 2)
            {
                node           = s.heap[++s.heap_len] = (max_code < 2 ? ++max_code : 0);
                tree[node * 2] = 1;
                s.depth[node]  = 0;
                s.opt_len--; if (stree != null)
                {
                    s.static_len -= stree[node * 2 + 1];
                }
                // node is 0 or 1 so it does not have extra bits
            }
            this.max_code = max_code;

            // The elements heap[heap_len/2+1 .. heap_len] are leaves of the tree,
            // establish sub-heaps of increasing lengths:

            for (n = s.heap_len / 2; n >= 1; n--)
            {
                s.pqdownheap(tree, n);
            }

            // Construct the Huffman tree by repeatedly combining the least two
            // frequent nodes.

            node = elems;                 // next internal node of the tree
            do
            {
                // n = node of least frequency
                n         = s.heap[1];
                s.heap[1] = s.heap[s.heap_len--];
                s.pqdownheap(tree, 1);
                m = s.heap[1];            // m = node of next least frequency

                s.heap[--s.heap_max] = n; // keep the nodes sorted by frequency
                s.heap[--s.heap_max] = m;

                // Create a new node father of n and m
                tree[node * 2]  = (short)(tree[n * 2] + tree[m * 2]);
                s.depth[node]   = (byte)(System.Math.Max(s.depth[n], s.depth[m]) + 1);
                tree[n * 2 + 1] = tree[m * 2 + 1] = (short)node;

                // and insert the new node in the heap
                s.heap[1] = node++;
                s.pqdownheap(tree, 1);
            }while(s.heap_len >= 2);

            s.heap[--s.heap_max] = s.heap[1];

            // At this point, the fields freq and dad are set. We can now
            // generate the bit lengths.

            gen_bitlen(s);

            // The field len is now set, we can generate the bit codes
            gen_codes(tree, max_code, s.bl_count);
        }
예제 #4
0
        internal void build_tree(Deflate s)
        {
            int num2;
            int num5;

            short[] tree      = this.dyn_tree;
            short[] numArray2 = this.stat_desc.static_tree;
            int     elems     = this.stat_desc.elems;
            int     num4      = -1;

            s.heap_len = 0;
            s.heap_max = 0x23d;
            for (num2 = 0; num2 < elems; num2++)
            {
                if (tree[num2 * 2] != 0)
                {
                    s.heap[++s.heap_len] = num4 = num2;
                    s.depth[num2]        = 0;
                }
                else
                {
                    tree[(num2 * 2) + 1] = 0;
                }
            }
            while (s.heap_len < 2)
            {
                num5           = s.heap[++s.heap_len] = (num4 >= 2) ? 0 : ++num4;
                tree[num5 * 2] = 1;
                s.depth[num5]  = 0;
                s.opt_len--;
                if (numArray2 != null)
                {
                    s.static_len -= numArray2[(num5 * 2) + 1];
                }
            }
            this.max_code = num4;
            num2          = s.heap_len / 2;
            while (num2 >= 1)
            {
                s.pqdownheap(tree, num2);
                num2--;
            }
            num5 = elems;
            do
            {
                num2      = s.heap[1];
                s.heap[1] = s.heap[s.heap_len--];
                s.pqdownheap(tree, 1);
                int index = s.heap[1];
                s.heap[--s.heap_max] = num2;
                s.heap[--s.heap_max] = index;
                tree[num5 * 2]       = (short)(tree[num2 * 2] + tree[index * 2]);
                s.depth[num5]        = (byte)(Math.Max(s.depth[num2], s.depth[index]) + 1);
                tree[(num2 * 2) + 1] = tree[(index * 2) + 1] = (short)num5;
                s.heap[1]            = num5++;
                s.pqdownheap(tree, 1);
            }while (s.heap_len >= 2);
            s.heap[--s.heap_max] = s.heap[1];
            this.gen_bitlen(s);
            gen_codes(tree, num4, s.bl_count);
        }