Solution
又看了半天的SAM，感覺這次看陳老師的講稿流暢多了。。
這道題的話，對 A 串建自動機，B串在上面匹配。
B 匹配到狀態(tài)u貢獻就是 |rightu|?(len?minu+1)=|rightu|?(len?maxparu) 。
再考慮匹配到這個點的話，必然在 parent 樹上它的祖先節(jié)點也是匹配的，又因為這些點的 maxv≤minu≤len 。所以 |rightv|?(maxv?minv+1) 是能夠全部計入方案的。
然后就是樹形DP了。
#include <bits/stdc++.h>
using namespace std;
const int N =
402020;
typedef long long ll;
inline char get(
void) {
static char buf[
100000], *S = buf, *T = buf;
if (S == T) {T = (S = buf) + fread(buf,
1,
100000, stdin);
if (S == T)
return EOF;}
return *S++;
}
template<
typename T>
inline void read(T &x) {
static char c; x =
0;
int sgn =
0;
for (c = get(); c <
'0' || c >
'9'; c = get())
if (c ==
'-') sgn =
1;
for (; c >=
'0' && c <=
'9'; c = get()) x = x *
10 + c -
'0';
if (sgn) x = -x;
}
inline void read(
char *ch) {
int len =
0;
char c;
for (c = get(); c <
'a' || c >
'z'; c = get());
for (; c >=
'a' && c <=
'z'; c = get()) ch[len++] = c;ch[len] =
0;
}
char a[N], b[N];
struct SAM {
int mx[N], par[N];
int ri[N], id[N], buc[N];ll g[N], f[N];
int to[N][
27];
int last, Tcnt, root;
int sta[N];
int top;SAM(
void) { Tcnt = root = last =
1; }
inline int Extend(
int c) {
int p = last, np = ++Tcnt;ri[np] =
1;mx[np] = mx[p] +
1;
for (; p && !to[p][c]; p = par[p])to[p][c] = np;
if (p) {
int q = to[p][c];
if (mx[q] != mx[p] +
1) {
int nq = ++Tcnt;mx[nq] = mx[p] +
1;
memcpy(to[nq], to[q],
sizeof to[q]);par[nq] = par[q];par[q] = par[np] = nq;
for (; p && to[p][c] == q; p = par[p])to[p][c] = nq;}
else {par[np] = q;}}
else {par[np] = root;}
return last = np;}
inline void Sort(
void) {
for (
int i =
1; i <= Tcnt; i++) ++buc[mx[i]];
for (
int i =
1; i <= Tcnt; i++) buc[i] += buc[i -
1];
for (
int i = Tcnt; i; i--) id[buc[mx[i]]--] = i;}
inline void dp(
void) {
for (
int i = Tcnt; i; i--) ri[par[id[i]]] += ri[id[i]];
for (
int i =
1; i <= Tcnt; i++) {
int pos = id[i];g[pos] = g[par[pos]] + (ll)ri[pos] * (mx[pos] - mx[par[pos]]);}}
inline int Insert(
char *begin,
char *end) {
for (
char* c = begin; c != end; c++)Extend(*c -
'a');}
inline ll Calc(
char *begin,
char *end) {ll ans =
0;
int p = root, len =
0;
for (
char *c = begin; c != end; c++) {
int x = *c -
'a';
if (to[p][x]) {++len; p = to[p][x];}
else {
while (p && !to[p][x]) p = par[p];
if (p) {len = mx[p] +
1; p = to[p][x];}
else {p = root; len =
0;}}
if (p != root) ans += g[par[p]] + (ll)(len - mx[par[p]]) * ri[p];}
return ans;}
inline void Debug(
int u =
1) {
for (
int i =
1; i <= top; i++)
putchar(sta[i] +
'a');
putchar(
'\n');
for (
int i =
0; i <
26; i++)
if (to[u][i]) {sta[++top] = i;Debug(to[u][i]);--top;}}
};
SAM S;
int main(
void) {freopen(
"1.in",
"r", stdin);freopen(
"1.out",
"w", stdout);read(a); read(b);S.Insert(a, a +
strlen(a));S.Sort(); S.dp();
cout << S.Calc(b, b +
strlen(b)) << endl;
return 0;
}
總結(jié)
以上是生活随笔為你收集整理的[后缀自动机][树形DP] BZOJ 4566: [Haoi2016]找相同字符的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。
如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。
