generated from bing/readnotes
210 lines
5.8 KiB
Markdown
210 lines
5.8 KiB
Markdown
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# 分支定界法
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分支定界法(branch and bound)是一种求解整数规划问题的最常用算法。这种方法不但可以求解纯整数规划,还可以求解混合整数规划问题。分支定界法是一种搜索与迭代的方法,选择不同的分支变量和子问题进行分支。
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通常,把全部可行解空间反复地分割为越来越小的子集,称为分支;并且对每个子集内的解集计算一个目标下界(对于最小值问题),这称为定界。在每次分枝后,凡是界限超出已知可行解集目标值的那些子集不再进一步分枝,这样,许多子集可不予考虑,这称剪枝。这就是分枝定界法的主要思路。
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***[百度百科](https://baike.baidu.com/item/%E5%88%86%E6%94%AF%E5%AE%9A%E7%95%8C%E6%B3%95)***
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## 问题示例
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最近在群里看到一个问题:
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给定m * n矩阵matrix,可以从任意位置开始,向上、向下、向左、向右移动,但要求下一个位置上的元素要大于当前元素。找出最长的递增路径长度。
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如图,矩阵元素`5`的上下左右分别是`2`,`8`,`4`,`6`
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![matrix](matrix.png)
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根据某算法大佬的指导,使用分支定界法解决此问题。
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将矩阵的每一个元素作为第一级分支A,对于每一个分支A相邻的上下左右四个元素,作为分支A的子分支B
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选定一个元素An, 对比An与AnBm的大小,当An小于AnBm时,找到AnBm对于的元素Ax,递归循环处理,直至找不到AnBm。每递归一次,路径长度+1,最后返回最大的路径长度。
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### 代码示例
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[代码文件](bnb.cpp)(*这是一份快速实现的代码,所以不一定是最优。*)
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```cpp
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#include <iostream>
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#include <vector>
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using namespace std;
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vector<vector<int>> randMatrix(int x, int y)
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{
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vector<vector<int>> matrix(x);
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for (int i = 0; i < x; i++)
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{
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matrix[i].resize(y);
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for (int j = 0; j < y; j++)
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{
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matrix[i][j] = rand() % 100;
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}
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}
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return matrix;
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}
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void display(vector<vector<int>> matrix, int x, int y)
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{
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for (int i = 0; i < x; i++)
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{
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for (int j = 0; j < y; j++)
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{
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cout << matrix[i][j] << " ";
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}
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cout << endl;
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}
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cout << endl;
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}
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vector<int> up(vector<vector<int>> matrix, int x, int y)
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{
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auto res = vector<int>(3) = {-1, -1, -1};
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if (x == 0)
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{
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return res;
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}
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res[0] = matrix[x - 1][y];
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res[1] = x - 1;
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res[2] = y;
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return res;
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}
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vector<int> down(vector<vector<int>> matrix, int x, int y)
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{
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auto res = vector<int>(3) = {-1, -1, -1};
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auto xlen = matrix.size();
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if (x == xlen - 1)
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{
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return res;
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}
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res[0] = matrix[x + 1][y];
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res[1] = x + 1;
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res[2] = y;
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return res;
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}
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vector<int> left(vector<vector<int>> matrix, int x, int y)
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{
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auto res = vector<int>(3) = {-1, -1, -1};
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if (y == 0)
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{
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return res;
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}
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res[0] = matrix[x][y - 1];
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res[1] = x;
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res[2] = y - 1;
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return res;
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}
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vector<int> right(vector<vector<int>> matrix, int x, int y)
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{
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auto res = vector<int>(3) = {-1, -1, -1};
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auto row = matrix[x];
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auto ylen = row.size();
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if (y == ylen - 1)
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{
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return res;
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}
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res[0] = matrix[x][y + 1];
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res[1] = x;
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res[2] = y + 1;
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return res;
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}
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vector<vector<vector<int>>> branch(vector<vector<int>> matrix, int x, int y)
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{
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auto branch = vector<vector<vector<int>>>(x * y);
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auto index = 0;
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for (int i = 0; i < x; i++)
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{
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for (int j = 0; j < y; j++)
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{
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branch[index].resize(7);
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branch[index][0].resize(1);
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branch[index][0][0] = matrix[i][j];
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branch[index][1].resize(1);
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branch[index][1][0] = i;
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branch[index][2].resize(1);
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branch[index][2][0] = j;
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branch[index][3] = up(matrix, i, j);
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branch[index][4] = down(matrix, i, j);
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branch[index][5] = left(matrix, i, j);
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branch[index][6] = right(matrix, i, j);
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index++;
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}
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}
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return branch;
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}
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int finditem(vector<vector<vector<int>>> branches, int tx, int ty, int plen = 0)
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{
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int ulen, dlen, llen, rlen;
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ulen = dlen = llen = rlen = plen;
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for (auto vi : branches)
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{
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if (vi[1][0] == tx && vi[2][0] == ty)
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{ //找到元素
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if (vi[3][0] > vi[0][0])
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{ //上
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cout << "up [" << vi[3][0] << "] is great than me [" << vi[0][0] << "]" << endl;
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ulen++;
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ulen = finditem(branches, vi[3][1], vi[3][2], ulen);
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}
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if (vi[4][0] > vi[0][0])
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{ //下
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cout << "down [" << vi[4][0] << "] is great than me [" << vi[0][0] << "]" << endl;
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dlen++;
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dlen = finditem(branches, vi[4][1], vi[4][2], dlen);
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}
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if (vi[5][0] > vi[0][0])
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{ //左
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cout << "left [" << vi[5][0] << "] is great than me [" << vi[0][0] << "]" << endl;
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llen++;
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llen = finditem(branches, vi[5][1], vi[5][2], llen);
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}
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if (vi[6][0] > vi[0][0])
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{ //右
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cout << "right [" << vi[6][0] << "] is great than me [" << vi[0][0] << "]" << endl;
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rlen++;
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rlen = finditem(branches, vi[6][1], vi[6][2], rlen);
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}
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}
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}
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plen = ulen;
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if (plen < dlen)
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{
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plen = dlen;
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}
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if (plen < llen)
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{
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plen = llen;
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}
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if (plen < rlen)
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{
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plen = rlen;
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}
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return plen;
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}
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int main()
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{
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const int x = 4;
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const int y = 4;
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auto matrix = randMatrix(x, y);
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display(matrix, x, y);
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auto branches = branch(matrix, x, y);
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// display(branches, x * y, 7);
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auto plen = finditem(branches, 0, 3);
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cout << matrix[0][3] << "\t" << plen << endl;
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return 0;
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}
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```
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