本文整理汇总了C++中SizeOptions::search方法的典型用法代码示例。如果您正苦于以下问题：C++ SizeOptions::search方法的具体用法？C++ SizeOptions::search怎么用？C++ SizeOptions::search使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类SizeOptions的用法示例。
在下文中一共展示了SizeOptions::search方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: costs
SetCovering(const SizeOptions& opt)
:
x(*this, num_alternatives, 0, 1),
z(*this, 0, 999999)
{
// costs per alternative
int _costs[] = {19, 16, 18, 13, 15, 19, 15, 17, 16, 15};
IntArgs costs(num_alternatives, _costs);
// the alternatives and the objects they contain
int _a[] = {
// 1 2 3 4 5 6 7 8 the objects
1,0,0,0,0,1,0,0, // alternative 1
0,1,0,0,0,1,0,1, // alternative 2
1,0,0,1,0,0,1,0, // alternative 3
0,1,1,0,1,0,0,0, // alternative 4
0,1,0,0,1,0,0,0, // alternative 5
0,1,1,0,0,0,0,0, // alternative 6
0,1,1,1,0,0,0,0, // alternative 7
0,0,0,1,1,0,0,1, // alternative 8
0,0,1,0,0,1,0,1, // alternative 9
1,0,0,0,0,1,1,0, // alternative 10
};
IntArgs a(num_alternatives*num_objects, _a);
for(int j = 0; j < num_objects; j++) {
IntVarArgs tmp;
for(int i = 0; i < num_alternatives; i++) {
tmp << expr(*this, x[i]*a[i*num_objects+j]);
}
if (opt.size() == 0) {
// set partition problem:
// objects must be covered _exactly_ once
rel(*this, sum(tmp) == 1);
} else {
// set covering problem
// all objects must be covered _at least_ once
rel(*this, sum(tmp) >= 1);
}
}
if (opt.search() == SEARCH_DFS) {
if (opt.size() == 0) {
rel(*this, z <= 49);
} else {
rel(*this, z <= 45);
}
}
linear(*this, costs, x, IRT_EQ, z);
branch(*this, x, INT_VAR_NONE(), INT_VAL_MIN());
}示例2: values
Investments(const SizeOptions& opt)
:
x(*this, num_projects, 0, 1),
total_persons(*this, 0, max_persons),
total_budget(*this, 0, max_budget),
total_projects(*this, 0, max_budget),
total_values(*this, 0, 99999) // large number
{
int _values[] = {600,400,100,150, 80,120,200,220, 90,380,290,130, 80,270,280};
IntArgs values(num_projects, _values);
int _budgets[] = {35,34,26,12,10,18,32,11,10,22,27,18,16,29,22};
IntArgs budgets(num_projects, _budgets);
int _personell[] = {5,3,4,2,2,2,4,1,1,5,3,2,2,4,3};
IntArgs personell(num_projects, _personell);
int num_requires = 5;
// 1-based
int _requires[] = {
3, 15,
4, 15,
8, 7,
13, 2,
14, 2
};
IntArgs requires(num_requires*2, _requires);
int num_not_with = 6;
// 1-based
int _not_with[] =
{
1, 10,
5, 6,
6, 5,
10, 1,
11, 15,
15, 11
};
IntArgs not_with(num_not_with*2, _not_with);
linear(*this, personell, x, IRT_EQ, total_persons);
linear(*this, budgets, x, IRT_EQ, total_budget);
rel(*this, total_projects==sum(x));
linear(*this, values, x, IRT_EQ, total_values);
// show all optimal solutions (there is exactly one)
if (opt.search() == SEARCH_DFS) {
rel(*this, total_values >= 2370);
}
//
// resource limits:
//
// total_budget <= max_budget
rel(*this, total_budget <= max_budget);
// total_persons <= max_persons
rel(*this, total_persons <= max_persons);
// total_projects <= max_projects
rel(*this, total_projects <= max_projects);
/*
Special requirements, using standard Integer Programming "tricks"
*/
// projects that require other projects
for(int i = 0; i < num_requires; i++) {
int r1 = requires[i*2+0] - 1;
int r2 = requires[i*2+1] - 1;
rel(*this, x[r1] - x[r2] <= 0);
}
// projects excluding other projects
for(int i = 0; i < num_not_with; i++) {
int n1 = not_with[i*2+0]-1;
int n2 = not_with[i*2+1]-1;
rel(*this, x[n1] + x[n2] <= 1);
}
//
// Branching.
//
branch(*this, x, INT_VAR_MAX_MAX(), INT_VAL_MAX());
}