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;
}
// '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);
}
}