思路:
这题有两种解法, 最标准的解法是树状数组套主席树(其实是套线段树), 可以做到在线解决, 缺点是写起来挺麻烦. (实际上用了离散化还是离线的)
同时题目可以离线解决, 意味着可以使用整体二分.
由于转为整体二分后变成单点修改区间查询, 所以使用树状数组.
代码:
(整体二分):1
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using namespace std;
const int Mn(100500);
void qrd(int& x) {
x = 0;
char c=getchar();
while(!isdigit(c)) {
c = getchar();
}
while(isdigit(c)) {
x = x*10 + c-'0';
c = getchar();
}
}
inline int lb(int x) { //lowbit
return x&(-x);
}
int n,m;
int sn;
int a[Mn],shf[Mn*2]; //原数组, 离散化数组
struct Query { //操作及询问
bool t; //是否修改
int x,y,z,id;
}q[Mn*3],tq[Mn*3]; //tq为辅助数组
int ans[Mn]; //答案
int c[Mn];
void add(int p,int x) { //BIT修改
for(;p<=n;p += lb(p)) {
c[p] += x;
}
}
int sum(int p) { //BIT前缀和
int ret(0);
for(;p;p -= lb(p)) {
ret += c[p];
}
return ret;
}
void solve(int l,int r,int ql,int qr) { //二分主函数
if(ql>qr) { //ql>qr说明无需处理, 直接返回
return;
}
if(l==r) {
for(int i(ql);i<=qr;++i) {
Query& nq(q[i]);
if(!nq.t) {
ans[nq.id] = shf[l]; //l==r对每个询问记录答案
}
}
return;
}
int mid((l+r)/2);
int pl(ql-1), pr(qr+1); //左右指针
for(int i(ql);i<=qr;++i) {
Query& nq(q[i]);
if(nq.t) {
if(nq.y<=shf[mid]) { //放入左边并修改
add(nq.x,nq.z);
tq[++pl] = nq;
} else { //放入右边
tq[--pr] = nq;
}
} else {
int sz = sum(nq.y) - sum(nq.x-1); //查询
if(sz >= nq.z) { //放入左边
tq[++pl] = nq;
} else {
nq.z -= sz; //注意减去左边区间贡献
tq[--pr] = nq;
}
}
}
for(int i(ql);i<=qr;++i) {
Query& nq(q[i]);
if(nq.t && nq.y<=shf[mid]) {
for(int p(nq.x);p<=n;p+=lb(p)) {
c[p] = 0; //清零
}
}
}
int cnt(ql-1);
for(int i(ql);i<=pl;++i) { //并入左区间
q[++cnt] = tq[i];
}
for(int i(qr);i>=pr;--i) { //并入右区间
q[++cnt] = tq[i];
}
solve(l,mid,ql,pl); //递归解决
solve(mid+1,r,pl+1,qr);
}
int main() {
qrd(n), qrd(m);
for(int i(1);i<=n;++i) {
qrd(a[i]);
q[i] = {true,i,a[i],1,0}; //将初始化视为添加操作
shf[i] = a[i];
}
sn = n;
int qrc(0),qc(n);
for(int i(n+1);i<=n+m;++i) {
char op;
cin >> op;
if(op=='Q') {
q[++qc].t = false;
qrd(q[qc].x), qrd(q[qc].y), qrd(q[qc].z);
q[qc].id = ++qrc;
} else {
int x,y;
qrd(x), qrd(y);
shf[++sn] = y;
q[++qc] = {true,x,a[x],-1,0}; //修改分为添加和删除
q[++qc] = {true,x,y,1,0};
a[x] = y;
}
}
sort(shf+1,shf+sn+1);
sn = unique(shf+1,shf+sn+1)-shf-1;
solve(1,sn,1,qc);
for(int i(1);i<=qrc;++i) {
printf("%d\n",ans[i]);
}
return 0;
}
(树套树):1
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using namespace std;
const int Mn(100500);
void qrd(int& x) {
x = 0;
char c=getchar();
while(!isdigit(c)) {
c = getchar();
}
while(isdigit(c)) {
x = x*10 + c-'0';
c = getchar();
}
}
inline int lb(int x) {
return x&(-x);
}
int a[Mn],shf[Mn*2];
struct Query {
int t,x,y,z;
}q[Mn];
struct Sgn { //线段树结点
int v;
Sgn *ls,*rs;
}*bitr[Mn];
Sgn *nil = new Sgn(); //空结点
void bud(int l,int r,Sgn* p) { //建空树
p->v = 0;
if(l==r) {
p->ls = p->rs = nil;
return;
}
int mid((l+r)/2);
p->ls = new Sgn();
p->rs = new Sgn();
bud(l,mid,p->ls);
bud(mid+1,r,p->rs);
}
void mdf(int l,int r,int mp,int mv,Sgn*& p) { //修改线段树
if(p==nil) {
p = new Sgn();
p->ls = p->rs = nil;
}
if(l==r) {
p->v += mv;
return;
}
int mid((l+r)/2);
if(mid>=mp) {
mdf(l,mid,mp,mv,p->ls);
} else {
mdf(mid+1,r,mp,mv,p->rs);
}
p->v = p->ls->v + p->rs->v;
}
Sgn *tmpq[2][Mn]; //查询根
int qcn[2]; //根数量
int qry(int l,int r,int k) { //查询
if(l==r) {
return shf[l];
}
//计算左儿子
int sum(0);
for(int i(1);i<=qcn[1];++i) {
sum += tmpq[1][i]->ls->v;
}
for(int i(1);i<=qcn[0];++i) {
sum -= tmpq[0][i]->ls->v;
}
int mid((l+r)/2);
if(k <= sum) {
for(int i(1);i<=qcn[1];++i) {
tmpq[1][i] = tmpq[1][i]->ls;
}
for(int i(1);i<=qcn[0];++i) {
tmpq[0][i] = tmpq[0][i]->ls;
}
return qry(l,mid,k);
} else {
for(int i(1);i<=qcn[1];++i) {
tmpq[1][i] = tmpq[1][i]->rs;
}
for(int i(1);i<=qcn[0];++i) {
tmpq[0][i] = tmpq[0][i]->rs;
}
return qry(mid+1,r,k-sum);
}
}
int main() {
nil->ls = nil->rs = nil; //初始化
int n,m;
qrd(n), qrd(m);
for(int i(1);i<=n;++i) {
qrd(a[i]);
shf[i] = a[i];
}
int sn(n);
for(int i(1);i<=m;++i) {
char op;
cin >> op;
if(op=='Q') {
q[i].t = 0;
qrd(q[i].x), qrd(q[i].y), qrd(q[i].z);
} else {
q[i].t = 1;
qrd(q[i].x), qrd(q[i].y);
shf[++sn] = q[i].y;
}
}
sort(shf+1,shf+sn+1);
sn = unique(shf+1,shf+sn+1)-shf-1;
for(int i(1);i<=n+1;++i) {
bitr[i] = nil;
}
for(int i(1);i<=n;++i) {
int t(i);
int mp = lower_bound(shf+1,shf+sn+1,a[i])-shf;
while(t<=n) {
mdf(1,sn,mp,1,bitr[t]);
t += lb(t);
}
}
for(int i(1);i<=m;++i) {
Query& d(q[i]);
if(d.t==0) {
if(d.y==1) {
printf("%d\n",a[1]);
continue;
}
qcn[0] = qcn[1] = 0;
d.x -= 1;
while(d.x) { //装入左端点相关根
tmpq[0][++qcn[0]] = bitr[d.x];
d.x -= lb(d.x);
}
while(d.y) { //装入右端点相关根
tmpq[1][++qcn[1]] = bitr[d.y];
d.y -= lb(d.y);
}
int ans = qry(1,sn,d.z);
printf("%d\n",ans);
} else {
int tmp(d.x);
int mp = lower_bound(shf+1,shf+sn+1,a[d.x])-shf;
while(tmp<=n) {
mdf(1,sn,mp,-1,bitr[tmp]); //嵌套修改
tmp += lb(tmp);
}
a[d.x] = d.y;
tmp = d.x;
mp = lower_bound(shf+1,shf+sn+1,d.y)-shf;
while(tmp<=n) {
mdf(1,sn,mp,1,bitr[tmp]);
tmp += lb(tmp);
}
}
}
return 0;
}