编程练习-动态规划（生产线问题）

line_A_cost是生产线A的代价;

line_B_cost是生产线B的代价

AB_cost是从生产线A转到生产线B的代价

BA_cost是从生产线B转到生产线A的代价

代码如下：

#include <stdio.h>
#include <vector>
void ProductionLine(std::vector<int>& line_A_cost, std::vector<int>& line_B_cost,
std::vector<int>& AB_cost, std::vector<int>& BA_cost) {
if (line_A_cost.size() != line_B_cost.size() ||
AB_cost.size() != BA_cost.size() ||
line_A_cost.size() - AB_cost.size() != 1 ||
line_A_cost.size() < 1) {
return;
}
std::vector<char> memo_A;
std::vector<char> memo_B;
std::vector<int> cost_A(1, line_A_cost[0]);
std::vector<int> cost_B(1, line_B_cost[0]);
int min_cost = 0;
for (int i = 1; i < line_A_cost.size(); ++i) {
if ((cost_A[i - 1] + line_A_cost[i]) < (cost_B[i - 1] + BA_cost[i - 1] + line_A_cost[i])) {
cost_A.push_back(cost_A[i - 1] + line_A_cost[i]);
memo_A.push_back('A');
printf("--%d cost A %d\n", i, cost_A[cost_A.size() - 1]);
} else {
cost_A.push_back(cost_B[i - 1] + BA_cost[i - 1] + line_A_cost[i]);
memo_A.push_back('B');
printf("%d cost A %d\n", i, cost_A[cost_A.size() - 1]);
}
if ((cost_B[i - 1] + line_B_cost[i]) < (cost_A[i - 1] + AB_cost[i - 1] + line_B_cost[i])) {
cost_B.push_back(cost_B[i - 1] + line_B_cost[i]);
memo_B.push_back('B');
printf("--%d cost B %d\n",i, cost_B[cost_B.size() - 1]);
} else {
cost_B.push_back(cost_A[i - 1] + AB_cost[i - 1] + line_B_cost[i]);
memo_B.push_back('A');
printf("%d cost B %d\n",i, cost_B[cost_B.size() - 1]);
}
}
printf("size=%zd", memo_A.size());

size_t last = cost_A.size() - 1;
std::vector<char>* current = NULL;
if (cost_A[last] > cost_B[last]) {
printf("B ");
current = &memo_B;
} else {
printf("A ");
current = &memo_A;
}
for (int j = current->size() - 1; j >= 0; --j) {
if ((*current)[j] == 'A') {
printf("A ");
current = &memo_A;
} else {
printf("B ");
current = &memo_B;
}
}
}
int main(int argc, char** argv) {
int A_array[] = {1, 3, 7, 2, 9};
std::vector<int> line_A_cost(A_array, A_array + sizeof(A_array) / sizeof(int));
int B_array[] = {2, 1, 3, 8, 10};
std::vector<int> line_B_cost(B_array, B_array + sizeof(B_array) / sizeof(int));
int AB_cost_array[] = {2, 3, 2, 4};
int BA_cost_array[] = {3, 1, 4, 2};
std::vector<int> AB_cost(AB_cost_array, AB_cost_array + sizeof(AB_cost_array) / sizeof(int));
std::vector<int> BA_cost(BA_cost_array, BA_cost_array + sizeof(BA_cost_array) / sizeof(int));
ProductionLine(line_A_cost, line_B_cost, AB_cost, BA_cost);
}

参考文献：

《算法导论》P193