本文整理汇总了C++中SegmentTree::initialize方法的典型用法代码示例。如果您正苦于以下问题:C++ SegmentTree::initialize方法的具体用法?C++ SegmentTree::initialize怎么用?C++ SegmentTree::initialize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SegmentTree
的用法示例。
在下文中一共展示了SegmentTree::initialize方法的5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: main
int main() {
int b, p, l, n, n_input;
while(scanf("%d %d %d %d", &b, &p, &l, &n) == 4) {
if(b == 0 and p == 0 and l == 0 and n == 0) break;
B = b;
mod = p;
n_input = n;
SegmentTree tree;
tree.initialize((int)l + 1);
char cm;
int i, v;
while(n_input--) {
assert(scanf(" %c %d %d", &cm, &i, &v) == 3);
//D(cm); D(i); D(v);
// responder al comando cm, i, v
if (cm == 'E') {
tree.update(i, v);
} else if (cm == 'H'){
int ans = tree.query(i, v) % mod;
printf("%d\n", ans);
}
}
puts("-");
}
return 0;
}
示例2: test
int
test() {
SegmentTree st;
st.initialize(14, 100);
st.insert(0, 45);
st.insert(1, 4);
st.insert(2, 5);
st.insert(3, 2);
st.insert(4, 99);
st.insert(5, 41);
st.insert(6, 45);
st.insert(7, 45);
st.insert(8, 51);
st.insert(9, 89);
st.insert(10, 1);
st.insert(11, 3);
st.insert(12, 5);
st.insert(13, 98);
vi_t v;
v.push_back(45);
v.push_back(4);
v.push_back(5);
v.push_back(2);
v.push_back(99);
v.push_back(41);
v.push_back(45);
v.push_back(45);
v.push_back(51);
v.push_back(89);
v.push_back(1);
v.push_back(3);
v.push_back(5);
v.push_back(98);
for (int i = 0; i < 14; ++i) {
for (int j = i; j < 14; ++j) {
pii_t one = st.query_min(i, j);
pii_t two = naive_query_min(v, i, j);
printf("query_min(%d, %d) == (%d, %d)\n", i, j, one.INDEX, two.INDEX);
assert(one.LENGTH == two.LENGTH);
}
}
return 0;
}
示例3: assert
// 'rin' is the input range set
//
// 'rout' is the output range set. It is guaranteed that 'rout'
// contains the mapping needed to re-contruct the least number of
// ranges needed to represent the complete range. Start with rout[0]
// and keep following the indexes till you reach some 'k' for which
// rout[k] == n.
//
// rout[i] contains the next index within rout (in sorted string
// order) where you need to jump to find the next range that completes
// the smallest number of ranges that cover the entire range.
//
// you will need to index into 'slcp' to locate the index of this
// range in the original string
//
void
compress_ranges(vi_t &rin, vi_t &rout) {
SegmentTree st;
rout.clear();
if (rin.empty()) {
return;
}
const int n = rin.size();
st.initialize(n + 1, n + 1);
rout.resize(n);
st.insert(n, 0);
for (int i = rin.size() - 1; i >= 0; --i) {
assert(rin[i] > 0);
int l = i + 1, r = std::min(n, i+rin[i]);
pii_t m = st.query_min(l, r);
rout[i] = m.INDEX;
st.insert(i, m.LENGTH + 1);
}
}
示例4: test
int
test() {
vui_t v;
v.push_back(45);
v.push_back(4);
v.push_back(5);
v.push_back(2);
v.push_back(99);
v.push_back(41);
v.push_back(45);
v.push_back(45);
v.push_back(51);
v.push_back(89);
v.push_back(1);
v.push_back(3);
v.push_back(5);
v.push_back(98);
for (int i = 0; i < 10; ++i) {
// printf("%d: %d\n", i, log2(i));
}
SegmentTree st;
st.initialize(v);
printf("Testing SegmentTree implementation\n");
printf("----------------------------------\n");
for (size_t i = 0; i < v.size(); ++i) {
for (size_t j = i; j < v.size(); ++j) {
pui_t one = st.query_max(i, j);
pui_t two = naive_query_max(v, i, j);
printf("query_max(%u, %u) == (%u, %u)\n", (uint_t)i, (uint_t)j, one.first, two.first);
assert_eq(one.first, two.first);
}
}
printf("\n");
return 0;
}
示例5: build_random_tree
void build_random_tree(int n = 100){
t.arr.resize(n);
for (int i=0; i<n; ++i) t.arr[i] = random() * (random() % 2 ? -1 : 1);
t.initialize();
}