本文主要是介绍钢条切割问题的解法（C/C++），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
（原题见算法导论·动态规划）
对长度为n的钢条进行切割，对应的切割长度和价格对应如下：
int cost[] = {0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30};
比如1对应价值1,10对应价值30。即相应的下标和值的对应。现求切割所得最大效益mx。
1.递归算法：
//g++ 编译通过int cut_rod(int *cost,int n)
{if(n == 0) return 0;int limit = MIN(n,10); //分割第一条的上限int mx = -1;for(int i = 1;i <= limit; ++i)mx = maxnum(mx,cost[i]+cut_rod(cost,n-i)); //取当前值于递归值的最大值return mx;
}由于对相同子问题的重复求解，T(n) = 2^n
2.递归标记数组算法：（自顶而下）（DFS）
//g++ 编译通过int mem_cut_rod(int *cost,int n,int *mem) //mem数组长度为n,所有元素须在其他函数中初始化为-1
{int mx;if (mem[n] >= 0) return mem[n]; //对于求过的问题，直接返回存储的值if (n == 0) mx = 0;else mx = -1;int limit = MIN(n,10);for(int i = 1;i <= limit; ++i)mx = maxnum(mx,cost[i]+mem_cut_rod(cost,n-i,mem)); //后面的内容和递归型是一样的mem[n] = mx; //储存计算出的新值return mx;
}3.逆拓扑序DP：（自底向上）
//g++ 编译通过int bottom_cut_rod(int *cost,int n)
{int mem[MEM_LEN+1]; //MEM_LEN = n，设置标记数组mem[0] = 0; //i,j将从1开始，这里收益是0for(int i = 1; i <= n; ++i) //从第一个问题开始求解{int mx = -1;int limit = MIN(i,10);for(int j = 1;j <= limit; ++j)mx = maxnum(mx,cost[j] + mem[i-j]); //求解最小的问题mem[i] = mx;}return mem[n];
}我们可以看到，2,3 的解法复杂度均为O(n^2)。
4.带解决方案的DFS：
typedef struct {string path; //方案路径bool memoried;int value;
} MEMORY;MEMORY *mem_pool;
string num_to_str(int num) {char buf[120];sprintf(buf, "%d", num);return string(buf);
}MEMORY DFS(int remain) {int select, limit = MIN(remain, COST_LEN), mx = -1, cur_cost;string cur_path, mx_path;if (mem_pool[remain - 1].memoried) {return mem_pool[remain - 1];}for (select = 1; select <= limit; ++select) {if (select == remain) {cur_cost = cost[remain];cur_path = num_to_str(remain);} else {MEMORY upper = DFS(remain - select);cur_cost = cost[select] + upper.value;cur_path = num_to_str(select) + ", " + upper.path;}if (cur_cost > mx) {mx = cur_cost;mx_path = cur_path;}}mem_pool[remain - 1].memoried = true;mem_pool[remain - 1].value = mx;mem_pool[remain - 1].path = mx_path;return mem_pool[remain - 1];
}int main() {int n, i;cin >> n;mem_pool = new MEMORY[n];if (!mem_pool) {return 1;}for (i = 0; i < n; ++i) {mem_pool[i].memoried = false;}MEMORY result = DFS(n);cout << result.value << endl;cout << result.path << endl;delete[] mem_pool;return 0;
}
最后我们附上一份c实现的优美代码：
//2015.6.2//copyright XJSoft#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>typedef struct {bool memoried;int value;
} MEMORY;int cost[] = {0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30};
MEMORY *mem_pool;#define COST_LEN 10
#define MIN(a,b) ((a)<(b)?(a):(b))
int maxnum(const int v1,const int v2)
{if (v1 > v2) return v1;else return v2;
}int DFS(int remain) {int select, limit = MIN(remain, COST_LEN), mx = -1, cur_cost;if (mem_pool[remain - 1].memoried) {return mem_pool[remain - 1].value;}for (select = 1; select <= limit; ++select) {if (select == remain) {cur_cost = cost[remain];} else {cur_cost = cost[select] + DFS(remain - select);}if (cur_cost > mx) {mx = cur_cost;}}mem_pool[remain - 1].memoried = true;mem_pool[remain - 1].value = mx;return mx;
}int DP(int n) {int remain, select, limit, mx, cur_cost;for (remain = 1; remain <= n; ++remain) {mx = -1;limit = MIN(remain, COST_LEN);for (select = 1; select <= limit; ++select) {if (select == remain) {cur_cost = cost[select];} else {cur_cost = cost[select] + mem_pool[remain - select - 1].value;}if (cur_cost > mx) {mx = cur_cost;}}mem_pool[remain - 1].value = mx;}return mem_pool[n - 1].value;
}int main() {int n, i;scanf("%d", &n);mem_pool = (MEMORY*)malloc(n * sizeof(MEMORY));if (!mem_pool) {printf("Mem error!\n");return 1;}for (i = 0; i < n; ++i) {mem_pool[i].memoried = false;}printf("%d\n", DP(n));free(mem_pool);return 0;
}
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