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# 分支定界法
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。
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。
***[百度百科](https://baike.baidu.com/item/%E5%88%86%E6%94%AF%E5%AE%9A%E7%95%8C%E6%B3%95)***
## 问题示例
最近在群里看到一个问题:
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。
如图,矩阵元素`5`的上下左右分别是`2`,`8`,`4`,`6`
![matrix](matrix.png)
根据某算法大佬的指导,使用分支定界法解决此问题。
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
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。
### 代码示例
[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。*)
```cpp
#include <iostream>
#include <vector>
using namespace std;
vector<vector<int>> randMatrix(int x, int y)
{
vector<vector<int>> matrix(x);
for (int i = 0; i < x; i++)
{
matrix[i].resize(y);
for (int j = 0; j < y; j++)
{
matrix[i][j] = rand() % 100;
}
}
return matrix;
}
void display(vector<vector<int>> matrix, int x, int y)
{
for (int i = 0; i < x; i++)
{
for (int j = 0; j < y; j++)
{
cout << matrix[i][j] << " ";
}
cout << endl;
}
cout << endl;
}
vector<int> up(vector<vector<int>> matrix, int x, int y)
{
auto res = vector<int>(3) = {-1, -1, -1};
if (x == 0)
{
return res;
}
res[0] = matrix[x - 1][y];
res[1] = x - 1;
res[2] = y;
return res;
}
vector<int> down(vector<vector<int>> matrix, int x, int y)
{
auto res = vector<int>(3) = {-1, -1, -1};
auto xlen = matrix.size();
if (x == xlen - 1)
{
return res;
}
res[0] = matrix[x + 1][y];
res[1] = x + 1;
res[2] = y;
return res;
}
vector<int> left(vector<vector<int>> matrix, int x, int y)
{
auto res = vector<int>(3) = {-1, -1, -1};
if (y == 0)
{
return res;
}
res[0] = matrix[x][y - 1];
res[1] = x;
res[2] = y - 1;
return res;
}
vector<int> right(vector<vector<int>> matrix, int x, int y)
{
auto res = vector<int>(3) = {-1, -1, -1};
auto row = matrix[x];
auto ylen = row.size();
if (y == ylen - 1)
{
return res;
}
res[0] = matrix[x][y + 1];
res[1] = x;
res[2] = y + 1;
return res;
}
vector<vector<vector<int>>> branch(vector<vector<int>> matrix, int x, int y)
{
auto branch = vector<vector<vector<int>>>(x * y);
auto index = 0;
for (int i = 0; i < x; i++)
{
for (int j = 0; j < y; j++)
{
branch[index].resize(7);
branch[index][0].resize(1);
branch[index][0][0] = matrix[i][j];
branch[index][1].resize(1);
branch[index][1][0] = i;
branch[index][2].resize(1);
branch[index][2][0] = j;
branch[index][3] = up(matrix, i, j);
branch[index][4] = down(matrix, i, j);
branch[index][5] = left(matrix, i, j);
branch[index][6] = right(matrix, i, j);
index++;
}
}
return branch;
}
int finditem(vector<vector<vector<int>>> branches, int tx, int ty, int plen = 0)
{
int ulen, dlen, llen, rlen;
ulen = dlen = llen = rlen = plen;
for (auto vi : branches)
{
if (vi[1][0] == tx && vi[2][0] == ty)
{ //找到元素
if (vi[3][0] > vi[0][0])
{ //上
cout << "up [" << vi[3][0] << "] is great than me [" << vi[0][0] << "]" << endl;
ulen++;
ulen = finditem(branches, vi[3][1], vi[3][2], ulen);
}
if (vi[4][0] > vi[0][0])
{ //下
cout << "down [" << vi[4][0] << "] is great than me [" << vi[0][0] << "]" << endl;
dlen++;
dlen = finditem(branches, vi[4][1], vi[4][2], dlen);
}
if (vi[5][0] > vi[0][0])
{ //左
cout << "left [" << vi[5][0] << "] is great than me [" << vi[0][0] << "]" << endl;
llen++;
llen = finditem(branches, vi[5][1], vi[5][2], llen);
}
if (vi[6][0] > vi[0][0])
{ //右
cout << "right [" << vi[6][0] << "] is great than me [" << vi[0][0] << "]" << endl;
rlen++;
rlen = finditem(branches, vi[6][1], vi[6][2], rlen);
}
}
}
plen = ulen;
if (plen < dlen)
{
plen = dlen;
}
if (plen < llen)
{
plen = llen;
}
if (plen < rlen)
{
plen = rlen;
}
return plen;
}
int main()
{
const int x = 4;
const int y = 4;
auto matrix = randMatrix(x, y);
display(matrix, x, y);
auto branches = branch(matrix, x, y);
// display(branches, x * y, 7);
auto plen = finditem(branches, 0, 3);
cout << matrix[0][3] << "\t" << plen << endl;
return 0;
}
```