day-4
效率挺低的。。
写了个概率dfs..esp应该调小点的 1e-13
有个分块树处理群运算(满足结合律)的板子 oiWiki.
//下标从0开始
SqrtTreeItem op(const SqrtTreeItem &a, const SqrtTreeItem &b);
inline int log2Up(int n) {
int res = 0;
while ((1 << res) < n) {
res++;
}
return res;
}
class SqrtTree {
private:
int n, lg, indexSz;
vector<SqrtTreeItem> v;
vector<int> clz, layers, onLayer;
vector<vector<SqrtTreeItem> > pref, suf, between;
inline void buildBlock(int layer, int l, int r) {
pref[layer][l] = v[l];
for (int i = l + 1; i < r; i++) {
pref[layer][i] = op(pref[layer][i - 1], v[i]);
}
suf[layer][r - 1] = v[r - 1];
for (int i = r - 2; i >= l; i--) {
suf[layer][i] = op(v[i], suf[layer][i + 1]);
}
}
inline void buildBetween(int layer, int lBound, int rBound, int betweenOffs) {
int bSzLog = (layers[layer] + 1) >> 1;
int bCntLog = layers[layer] >> 1;
int bSz = 1 << bSzLog;
int bCnt = (rBound - lBound + bSz - 1) >> bSzLog;
for (int i = 0; i < bCnt; i++) {
SqrtTreeItem ans;
for (int j = i; j < bCnt; j++) {
SqrtTreeItem add = suf[layer][lBound + (j << bSzLog)];
ans = (i == j) ? add : op(ans, add);
between[layer - 1][betweenOffs + lBound + (i << bCntLog) + j] = ans;
}
}
}
inline void buildBetweenZero() {
int bSzLog = (lg + 1) >> 1;
for (int i = 0; i < indexSz; i++) {
v[n + i] = suf[0][i << bSzLog];
}
build(1, n, n + indexSz, (1 << lg) - n);
}
inline void updateBetweenZero(int bid) {
int bSzLog = (lg + 1) >> 1;
v[n + bid] = suf[0][bid << bSzLog];
update(1, n, n + indexSz, (1 << lg) - n, n + bid);
}
void build(int layer, int lBound, int rBound, int betweenOffs) {
if (layer >= (int)layers.size()) {
return;
}
int bSz = 1 << ((layers[layer] + 1) >> 1);
for (int l = lBound; l < rBound; l += bSz) {
int r = min(l + bSz, rBound);
buildBlock(layer, l, r);
build(layer + 1, l, r, betweenOffs);
}
if (layer == 0) {
buildBetweenZero();
} else {
buildBetween(layer, lBound, rBound, betweenOffs);
}
}
void update(int layer, int lBound, int rBound, int betweenOffs, int x) {
if (layer >= (int)layers.size()) {
return;
}
int bSzLog = (layers[layer] + 1) >> 1;
int bSz = 1 << bSzLog;
int blockIdx = (x - lBound) >> bSzLog;
int l = lBound + (blockIdx << bSzLog);
int r = min(l + bSz, rBound);
buildBlock(layer, l, r);
if (layer == 0) {
updateBetweenZero(blockIdx);
} else {
buildBetween(layer, lBound, rBound, betweenOffs);
}
update(layer + 1, l, r, betweenOffs, x);
}
inline SqrtTreeItem query(int l, int r, int betweenOffs, int base) {
if (l == r) {
return v[l];
}
if (l + 1 == r) {
return op(v[l], v[r]);
}
int layer = onLayer[clz[(l - base) ^ (r - base)]];
int bSzLog = (layers[layer] + 1) >> 1;
int bCntLog = layers[layer] >> 1;
int lBound = (((l - base) >> layers[layer]) << layers[layer]) + base;
int lBlock = ((l - lBound) >> bSzLog) + 1;
int rBlock = ((r - lBound) >> bSzLog) - 1;
SqrtTreeItem ans = suf[layer][l];
if (lBlock <= rBlock) {
SqrtTreeItem add =
(layer == 0) ? (query(n + lBlock, n + rBlock, (1 << lg) - n, n))
: (between[layer - 1][betweenOffs + lBound +
(lBlock << bCntLog) + rBlock]);
ans = op(ans, add);
}
ans = op(ans, pref[layer][r]);
return ans;
}
public:
inline SqrtTreeItem query(int l, int r) { return query(l, r, 0, 0); }
inline void update(int x, const SqrtTreeItem &item) {
v[x] = item;
update(0, 0, n, 0, x);
}
SqrtTree(const vector<SqrtTreeItem> &a)
: n((int)a.size()), lg(log2Up(n)), v(a), clz(1 << lg), onLayer(lg + 1) {
clz[0] = 0;
for (int i = 1; i < (int)clz.size(); i++) {
clz[i] = clz[i >> 1] + 1;
}
int tlg = lg;
while (tlg > 1) {
onLayer[tlg] = (int)layers.size();
layers.push_back(tlg);
tlg = (tlg + 1) >> 1;
}
for (int i = lg - 1; i >= 0; i--) {
onLayer[i] = max(onLayer[i], onLayer[i + 1]);
}
int betweenLayers = max(0, (int)layers.size() - 1);
int bSzLog = (lg + 1) >> 1;
int bSz = 1 << bSzLog;
indexSz = (n + bSz - 1) >> bSzLog;
v.resize(n + indexSz);
pref.assign(layers.size(), vector<SqrtTreeItem>(n + indexSz));
suf.assign(layers.size(), vector<SqrtTreeItem>(n + indexSz));
between.assign(betweenLayers, vector<SqrtTreeItem>((1 << lg) + bSz));
build(0, 0, n, 0);
}
};
dp 好难啊啊。。。。
组合数学也。。
我看见 你