題目大意
一個字符串長度爲10^5,由’e,a,s,y’四種字符組成,給定一個閉區間[l,r],問區間之內最多有多少個互不重疊的子序列easy。
例如對於eeaseyaesasyy,只有兩個easy。
經典的倍增查詢區間值問題。
對於一個字符串,一定是從左端點開始貪心地找子序列是最佳的方案,也就是需要固定左端點。那麼只需要維護每個左端點開始的答案就可以了。而實際要求的子序列就是多個easy的重複,就是說明白序列是什麼樣子的,因而我可以知道序列的第
區間查詢也可以用線段樹,合併要仔細地討論一下。
//倍增
#include <cstdio>
#include <cstring>
char s[100005];
int m, n, pe, pa, ps, py, dep;
int pre[100005], ed[100005];
int fa[100005][20];
int main() {
scanf("%s", s);
scanf("%d", &m);
n = strlen(s);
for (int i = 0; i < n; ++i) {
if (s[i] == 'e') {
pre[i + 1] = py;
pe = i + 1;
}
if (s[i] == 'a') {
pre[i + 1] = pe;
pa = i + 1;
}
if (s[i] == 's') {
pre[i + 1] = pa;
ps = i + 1;
}
if (s[i] == 'y') {
pre[i + 1] = ps;
py = i + 1;
}
ed[i + 1] = py;
}
for (int i = 1; i <= n; ++i) fa[i][0] = pre[i];
for (dep = 1; 1 << dep < n; ++dep);
for (int i = 1; i < dep; ++i)
for (int j = 1; j <= n; ++j)
fa[j][i] = fa[fa[j][i - 1]][i - 1];
for (; m--;) {
int l, r;
scanf("%d%d", &l, &r);
int x = ed[r];
int res = 1;
for (int i = dep - 1; i >= 0; --i) {
if (fa[x][i] >= l) {
x = fa[x][i];
res += 1 << i;
}
}
printf("%d\n", res / 4);
}
return 0;
}
//線段樹
#include <iostream>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <utility>
#include <algorithm>
#include <iomanip>
#define sqr(x) ((x)*(x))
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fi first
#define se second
#define mp make_pair
#define eps 1e-9
using namespace std;
typedef long long LL;
inline int read() {
static char ch;
bool sgn = false;
while (ch = getchar(), ch < '0' || ch > '9') if (ch == '-') sgn = true;
int res = ch - 48;
while (ch = getchar(), ch >= '0' && ch <= '9') res = res * 10 + ch - 48;
return sgn ? -res : res;
}
const int N=1e5+5;
char s[N];
int n,m,a[N],nxt[N];
int getnum(char c) {
if(c=='e') return 0;
if(c=='a') return 1;
if(c=='s') return 2;
if(c=='y') return 3;
}
struct Node {
int f[4],en[4];
Node(){
memset(f,0,sizeof(f));
memset(en,0,sizeof(en));
}
}tree[N<<3];
Node merge(Node a,Node b) {
Node res;
for(int i=0;i<4;i++) {
if(a.f[i]==0) {
res.f[i]=b.f[i];
res.en[i]=b.en[i];
continue;
}
if(a.en[i]==3) {
if(b.f[0]==0) {
res.f[i]=a.f[i];
res.en[i]=a.en[i];
} else {
res.f[i]=a.f[i]+b.f[0];
res.en[i]=b.en[0];
}
} else {
if(b.f[a.en[i]+1]==0) {
res.f[i]=a.f[i];
res.en[i]=a.en[i];
} else {
res.f[i]=a.f[i]+b.f[a.en[i]+1]-1;
res.en[i]=b.en[a.en[i]+1];
}
}
}
return res;
}
void build(int k,int l,int r) {
if(l==r) {
tree[k].f[a[l]]=1;
tree[k].en[a[l]]=a[l];
return;
}
int mid=l+r>>1;
build(k<<1,l,mid); build(k<<1|1,mid+1,r);
tree[k]=merge(tree[k<<1],tree[k<<1|1]);
}
Node Query(int k,int l,int r,int p,int q) {
if(l>=p&&r<=q) return tree[k];
int mid=l+r>>1;
if(mid>=q) return Query(k<<1,l,mid,p,q);
if(mid+1<=p) return Query(k<<1|1,mid+1,r,p,q);
return merge(Query(k<<1,l,mid,p,q),Query(k<<1|1,mid+1,r,p,q));
}
int main() {
scanf("%s",s+1);
n=strlen(s+1);
for(int i=0;i<4;i++) nxt[i]=i+1; nxt[3]=0;
for(int i=1;i<=n;i++) a[i]=getnum(s[i]);
build(1,1,n);
m=read();
while(m--) {
int l=read(),r=read();
Node ans=Query(1,1,n,l,r);
//cout<< ans.f[0]<<"/"<<endl;
printf("%d\n",ans.f[0]-(ans.en[0]!=3&&ans.f[0]>0));
}
}