新科技：整体二分
它能解决的典型问题：带修改区间第(k)大
大概的做法是这样的：我们一次二分一个值(mid)，然后依据操作的答案与(mid)的大小关系把操作分别划到两边，然后递归下去。也就是相当于二分的是所有询问的答案
感觉其实这个跟在权值线段树上二分一个效果，只是用离线的方式替代掉了那一层权值线段树而已
计算可得复杂度为(O(nlog^2n))（由主定理，(T(n)=2T(n/2)+O(nlogn)=O(nlog^2n))）
拿线段树或者树状数组维护都行
板子题是这一道K大数查询
直接上代码了，很好写，跟(cdq)分治写起来很像
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline
#define N 100000
#define M 100000
int n, m, qtot, ans[M+5];
struct Query {
int type;
int a, b, id;
ll c;
}q[M+5], tl[M+5], tr[M+5];
struct SegmentTree {
ll sumv[4*N+5], addv[4*N+5];
int lson(int o) { return (o<<1); }
int rson(int o) { return (o<<1|1); }
void init() {
memset(sumv, 0, sizeof sumv);
memset(addv, 0, sizeof addv);
}
void pushup(int o) {
sumv[o] = sumv[lson(o)]+sumv[rson(o)];
}
void pushdown(int o, int l, int r) {
if(addv[o]) {
int mid = (l+r)>>1;
sumv[lson(o)] += (mid-l+1)*addv[o], addv[lson(o)] += addv[o];
sumv[rson(o)] += (r-mid)*addv[o], addv[rson(o)] += addv[o];
addv[o] = 0;
}
}
void add(int o, int l, int r, int L, int R, ll k) {
if(L <= l && r <= R) {
sumv[o] += (r-l+1)*k, addv[o] += k;
return ;
}
pushdown(o, l, r);
int mid = (l+r)>>1;
if(L <= mid) add(lson(o), l, mid, L, R, k);
if(R > mid) add(rson(o), mid+1, r, L, R, k);
pushup(o);
}
ll query(int o, int l, int r, int L, int R) {
if(L <= l && r <= R) return sumv[o];
pushdown(o, l, r);
int mid = (l+r)>>1;
ll ret = 0;
if(L <= mid) ret += query(lson(o), l, mid, L, R);
if(R > mid) ret += query(rson(o), mid+1, r, L, R);
return ret;
}
}ST;
void solve(int l, int r, int L, int R) {
if(l > r) return ;
if(L == R) {
for(int i = l; i <= r; ++i) if(q[i].type == 2) ans[q[i].id] = L;
return ;
}
int mid = L+(R-L)/2;
int tot1 = 0, tot2 = 0;
for(int i = l; i <= r; ++i) {
if(q[i].type == 1) {
if(q[i].c > mid) ST.add(1, 1, n, q[i].a, q[i].b, 1), tr[++tot2] = q[i];
else tl[++tot1] = q[i];
}
else {
ll v = ST.query(1, 1, n, q[i].a, q[i].b);
if(v >= q[i].c) tr[++tot2] = q[i];
else q[i].c -= v, tl[++tot1] = q[i];
}
}
for(int i = 1; i <= tot1; ++i) q[l+i-1] = tl[i];
for(int i = 1; i <= tot2; ++i) {
q[l+tot1+i-1] = tr[i];
if(tr[i].type == 1) ST.add(1, 1, n, tr[i].a, tr[i].b, -1);
}
solve(l, l+tot1-1, L, mid), solve(l+tot1, r, mid+1, R);
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) {
scanf("%d%d%d%lld", &q[i].type, &q[i].a, &q[i].b, &q[i].c);
if(q[i].type == 2) q[i].id = ++qtot;
}
solve(1, m, -n, n);
for(int i = 1; i <= qtot; ++i) printf("%d
", ans[i]);
return 0;
}
Another One?
还有一道比较板的题 MET-Meteors，这题可能需要稍微卡卡常数才能过，直接把代码扔过来了
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline
#define N 300000
#define M 300000
#define K 300000
inline char Getchar(){
static char buf[512], *p1=buf, *p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,512,stdin),p1==p2)?EOF:*p1++;
}
inline int read() {
int s = 0; char c = Getchar();
while(c < '0' || c > '9') c = Getchar();
while(c >= '0' && c <= '9') s = s*10+c-'0', c = Getchar();
return s;
}
int n, m, k, tot;
int cnt[N+5], head[N+5], want[N+5], belong[M+5], ans[N+5];
struct Opt {
bool type;
int l, r, a;
int time;
int *pos;
}q[M+N+5], swp[M+N+5];
struct BIT {
ll b[M+5];
int lowbit(int x) {
return x&-x;
}
void update(int x, ll k) {
if(!x) return ;
while(x <= m) {
b[x] += k;
x += lowbit(x);
}
}
void add(int l, int r, ll k) {
if(l <= r) update(l, k), update(r+1, -k);
else update(l, k), update(m+1, -k), update(1, k), update(r+1, -k);
}
ll query(int x) {
ll ret = 0;
while(x) {
ret += b[x];
x -= lowbit(x);
}
return ret;
}
}bit;
ll getsum(Opt &q) {
ll ret = 0;
for(int i = 0; i < cnt[q.l]; ++i) {
ll v = bit.query(q.pos[i]);
if(v >= IINF-ret) ret = INF;
ret += v;
}
return ret;
}
void solve(int l, int r, int L, int R) {
if(l > r) return ;
if(L == R) {
for(int i = l; i <= r; ++i) if(q[i].type == 1) ans[q[i].l] = L;
return ;
}
int i, mid = L+(R-L)/2, tot1 = 0, tot2 = r-l+2;
for(i = l; i <= r; ++i) {
if(q[i].type == 0) {
if(q[i].time <= mid) bit.add(q[i].l, q[i].r, q[i].a), swp[++tot1] = q[i];
else swp[--tot2] = q[i];
}
else {
ll v = getsum(q[i]);
if(v >= q[i].a) swp[++tot1] = q[i];
else q[i].a -= v, swp[--tot2] = q[i];
}
}
for(i = 1; i <= tot1; ++i) {
if(swp[i].type == 0) bit.add(swp[i].l, swp[i].r, -swp[i].a);
q[i+l-1] = swp[i];
}
for(i = r-l+1; i >= tot2; --i) q[r+1-i+tot1] = swp[i];
solve(l, l+tot1-1, L, mid), solve(l+tot1, r, mid+1, R);
}
int main() {
n = read(), m = read();
for(int i = 1; i <= m; ++i) belong[i] = read(), cnt[belong[i]]++;
for(int i = 1; i <= n; ++i) want[i] = read();
k = read();
for(int i = 1, l, r, a; i <= k; ++i) {
l = read(), r = read(), a = read();
q[++tot] = {0, l, r, a, i};
}
for(int i = 1; i <= n; ++i) {
q[tot+i] = {1, i, i, want[i]};
q[tot+i].pos = new int[cnt[i]];
}
for(int i = 1; i <= m; ++i) q[tot+belong[i]].pos[head[belong[i]]++] = i;
tot += n;
solve(1, tot, 1, k+1);
for(int i = 1; i <= n; ++i) {
if(ans[i] == k+1) printf("NIE
");
else printf("%d
", ans[i]);
}
return 0;
}